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Summer Assignment ACP Pre-calculus Do as many problems as you are able to do P1 Algebraic expressions, mathematical models, and real numbers Page 15-17 Questions 11-39 every other odd, 85-101 odd P2 Exponents and scientific notation Page 28 Questions 5- 5 by 5’s P4 Polynomials Page 53 Questions 1-81 odd P7 Equations Linear, absolute value, quadratic Page 981 Questions 1-15 odd, 43-54 all, 55-109 odd Prerequisites: Fundamental Concepts of Algebra P What can algebra possibly have to tell me about • the skyrocketing cost of a college education? • my workouts? • apologizing to a friend for a blunder I committed? • the meaning of the national debt that exceeds $9 trillion? • time dilation on a futuristic high-speed journey to a nearby star? • the widening imbalance between numbers of women and men on college campuses? This chapter reviews fundamental concepts of algebra that are prerequisites for the study of precalculus. Throughout the chapter, you will see how the special language of algebra describes your world. Here’s where you’ll find these applications: • College costs: Section P.1, Example 2; Exercise Set P.1, Exercises 131–132 • Workouts: Exercise Set P.1, Exercises 129–130 • Apologizing: Essay on page 15 • The national debt: Section P.2, Example 6 • Time dilation: Essay on page 42 • College gender imbalance: Chapter P Test, Exercise 32. 1 2 Chapter P Prerequisites: Fundamental Concepts of Algebra Section P.1 Objectives Algebraic Expressions, Mathematical Models, and Real Numbers H � Evaluate algebraic ow would your lifestyle change if a gallon of gas cost $9.15? Or if the price of a staple such as milk was $15? That’s how much those products would cost if their prices had increased at the same rate college tuition has increased since 1980. (Source: Center for College Affordability and Productivity) In this section, you will learn how the special language of algebra describes your world, including the skyrocketing cost of a college education. expressions. � Use mathematical models. � Find the intersection of two sets. � Find the union of two sets. � Recognize subsets of the real numbers. � Use inequality symbols. � Evaluate absolute value. � Use absolute value to express distance. � Identify properties of the real numbers. 쐅 Simplify algebraic expressions. Algebraic Expressions Algebra uses letters, such as x and y, to represent numbers. If a letter is used to represent various numbers, it is called a variable. For example, imagine that you are basking in the sun on the beach. We can let x represent the number of minutes that you can stay in the sun without burning with no sunscreen. With a number 6 sunscreen, exposure time without burning is six times as long, or 6 times x. This can be written 6 # x, but it is usually expressed as 6x. Placing a number and a letter next to one another indicates multiplication. Notice that 6x combines the number 6 and the variable x using the operation of multiplication. A combination of variables and numbers using the operations of addition, subtraction, multiplication, or division, as well as powers or roots, is called an algebraic expression. Here are some examples of algebraic expressions: x + 6, x - 6, 6x, x , 3x + 5, x2 - 3, 1x + 7. 6 Many algebraic expressions involve exponents. For example, the algebraic expression 17x 2 + 261x + 3257 approximates the average cost of tuition and fees at public U.S. colleges for the school year ending x years after 2000. The expression x2 means x # x, and is read “x to the second power” or “x squared.” The exponent, 2, indicates that the base, x, appears as a factor two times. Exponential Notation If n is a counting number (1, 2, 3, and so on), Exponent or Power bn= b ⴢ b ⴢ b ⴢ . . . ⴢ b. Base b appears as a factor n times. bn is read “the nth power of b” or “b to the nth power.” Thus, the nth power of b is defined as the product of n factors of b. The expression bn is called an exponential expression. Furthermore, b1 = b. For example, 82 = 8 # 8 = 64, 53 = 5 # 5 # 5 = 125, and 2 4 = 2 # 2 # 2 # 2 = 16. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 3 � Evaluate algebraic expressions. Evaluating Algebraic Expressions Evaluating an algebraic expression means to find the value of the expression for a given value of the variable. Many algebraic expressions involve more than one operation. Evaluating an algebraic expression without a calculator involves carefully applying the following order of operations agreement: The Order of Operations Agreement 1. Perform operations within the innermost parentheses and work outward. If the algebraic expression involves a fraction, treat the numerator and the denominator as if they were each enclosed in parentheses. 2. Evaluate all exponential expressions. 3. Perform multiplications and divisions as they occur, working from left to right. 4. Perform additions and subtractions as they occur, working from left to right. Evaluating an Algebraic Expression EXAMPLE 1 Evaluate 7 + 51x - 423 for x = 6. Solution 7 + 51x - 423 = 7 + 516 - 423 = 7 + 51223 = 7 + 5182 = 7 + 40 = 47 Check Point � Use mathematical models. 1 Replace x with 6. First work inside parentheses: 6 - 4 = 2. Evaluate the exponential expression: 23 = 2 # 2 # 2 = 8. Multiply: 5182 = 40. Add. Evaluate 8 + 61x - 322 for x = 13. Formulas and Mathematical Models An equation is formed when an equal sign is placed between two algebraic expressions. One aim of algebra is to provide a compact, symbolic description of the world. These descriptions involve the use of formulas. A formula is an equation that uses variables to express a relationship between two or more quantities. Here are two examples of formulas related to heart rate and exercise. Couch-Potato Exercise 1 H= (220-a) 5 Heart rate, in beats per minute, is 1 of 5 the difference between 220 and your age. Working It 9 H= (220-a) 10 Heart rate, in beats per minute, is 9 of 10 the difference between 220 and your age. 4 Chapter P Prerequisites: Fundamental Concepts of Algebra The process of finding formulas to describe real-world phenomena is called mathematical modeling. Such formulas, together with the meaning assigned to the variables, are called mathematical models. We often say that these formulas model, or describe, the relationships among the variables. EXAMPLE 2 Modeling the Cost of Attending a Public College The bar graph in Figure P.1 shows the average cost of tuition and fees for public four-year colleges, adjusted for inflation.The formula T = 17x 2 + 261x + 3257 models the average cost of tuition and fees, T, for public U.S. colleges for the school year ending x years after 2000. a. Use the formula to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2007. b. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1? Average Cost of Tuition and Fees at Public Four-Year United States Colleges $6200 5836 $5800 5491 Tuition and Fees $5400 5132 $5000 4694 $4600 4081 $4200 3725 $3800 $3400 3362 3487 $3000 2000 2001 2002 2003 2004 2005 2006 2007 Ending Year in the School Year Figure P.1 Source: The College Board Solution a. Because 2007 is 7 years after 2000, we substitute 7 for x in the given formula. Then we use the order of operations to find T, the average cost of tuition and fees for the school year ending in 2007. T = 17x2 + 261x + 3257 This is the given mathematical model. T = 171722 + 261172 + 3257 Replace each occurrence of x with 7. T = 171492 + 261172 + 3257 Evaluate the exponential expression 72 = 7 # 7 = 49. T = 833 + 1827 + 3257 Multiply from left to right: 171492 = 833 and 261172 = 1827. T = 5917 Add. The formula indicates that for the school year ending in 2007, the average cost of tuition and fees at public U.S. colleges was $5917. b. Figure P.1 shows that the average cost of tuition and fees for the school year ending in 2007 was $5836. The cost obtained from the formula, $5917, overestimates the actual data value by $5917 - $5836, or by $81. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 5 2 Assuming trends indicated by the data in Figure P.1 continue, use the formula T = 17x2 + 261x + 3257, described in Example 2, to project the average cost of tuition and fees at public U.S. colleges for the school year ending in 2010. Check Point Sometimes a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense. In these cases, we say that model breakdown has occurred. For example, it is not likely that the formula in Example 2 would give a good estimate of tuition and fees in 2050 because it is too far in the future. Thus, model breakdown would occur. Sets Before we describe the set of real numbers, let’s be sure you are familiar with some basic ideas about sets. A set is a collection of objects whose contents can be clearly determined. The objects in a set are called the elements of the set. For example, the set of numbers used for counting can be represented by 51, 2, 3, 4, 5, Á 6. Study Tip Grouping symbols such as parentheses, ( ), and square brackets, [ ], are not used to represent sets. Only commas are used to separate the elements of a set. Separators such as colons or semicolons are not used. The braces, 5 6, indicate that we are representing a set. This form of representation, called the roster method, uses commas to separate the elements of the set. The symbol consisting of three dots after the 5, called an ellipsis, indicates that there is no final element and that the listing goes on forever. A set can also be written in set-builder notation. In this notation, the elements of the set are described, but not listed. Here is an example: {x|x is a counting number less than 6}. The set of all x such that x is a counting number less than 6. The same set written using the roster method is � Find the intersection of two sets. 51, 2, 3, 4, 56. If A and B are sets, we can form a new set consisting of all elements that are in both A and B. This set is called the intersection of the two sets. Definition of the Intersection of Sets A B The intersection of sets A and B, written A ¨ B, is the set of elements common to both set A and set B. This definition can be expressed in set-builder notation as follows: A ¨ B = 5x ƒ x is an element of A AND x is an element of B6. A傽B Figure P.2 Picturing the intersection of two sets Figure P.2 shows a useful way of picturing the intersection of sets A and B. The figure indicates that A ¨ B contains those elements that belong to both A and B at the same time. Finding the Intersection of Two Sets EXAMPLE 3 Find the intersection: 57, 8, 9, 10, 116 ¨ 56, 8, 10, 126. Solution The elements common to 57, 8, 9, 10, 116 and 56, 8, 10, 126 are 8 and 10. Thus, 57, 8, 9, 10, 116 ¨ 56, 8, 10, 126 = 58, 106. Check Point 3 Find the intersection: 53, 4, 5, 6, 76 ¨ 53, 7, 8, 96. 6 Chapter P Prerequisites: Fundamental Concepts of Algebra If a set has no elements, it is called the empty set, or the null set, and is represented by the symbol ¤ (the Greek letter phi). Here is an example that shows how the empty set can result when finding the intersection of two sets: {2, 4, 6} 艚 {3, 5, 7}=¤. These sets have no common elements. � Find the union of two sets. Their intersection has no elements and is the empty set. Another set that we can form from sets A and B consists of elements that are in A or B or in both sets. This set is called the union of the two sets. Definition of the Union of Sets A B A艛B The union of sets A and B, written A ´ B, is the set of elements that are members of set A or of set B or of both sets. This definition can be expressed in set-builder notation as follows: A ´ B = 5x ƒ x is an element of A OR x is an element of B6. Figure P.3 Picturing the union of two sets Study Tip When finding the union of two sets, do not list twice any elements that appear in both sets. Figure P.3 shows a useful way of picturing the union of sets A and B. The figure indicates that A ´ B is formed by joining the sets together. We can find the union of set A and set B by listing the elements of set A. Then, we include any elements of set B that have not already been listed. Enclose all elements that are listed with braces. This shows that the union of two sets is also a set. Finding the Union of Two Sets EXAMPLE 4 Find the union: 57, 8, 9, 10, 116 ´ 56, 8, 10, 126. Solution To find 57, 8, 9, 10, 116 ´ 56, 8, 10, 126, start by listing all the elements from the first set, namely 7, 8, 9, 10, and 11. Now list all the elements from the second set that are not in the first set, namely 6 and 12. The union is the set consisting of all these elements. Thus, 57, 8, 9, 10, 116 ´ 56, 8, 10, 126 = 56, 7, 8, 9, 10, 11, 126. Check Point � Recognize subsets of the real numbers. 4 Find the union: 53, 4, 5, 6, 76 ´ 53, 7, 8, 96. The Set of Real Numbers The sets that make up the real numbers are summarized in Table P.1. We refer to these sets as subsets of the real numbers, meaning that all elements in each subset are also elements in the set of real numbers. Notice the use of the symbol L in the examples of irrational numbers. The symbol means “is approximately equal to.” Thus, 22 L 1.414214. Technology A calculator with a square root key gives a decimal approximation for 22, not the exact value. We can verify that this is only an approximation by multiplying 1.414214 by itself. The product is very close to, but not exactly, 2: 1.414214 * 1.414214 = 2.000001237796. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 7 Table P.1 Important Subsets of the Real Numbers Name Description Natural numbers ⺞ {1, 2, 3, 4, 5, . . . } These are the numbers that we use for counting. 2, 3, 5, 17 Whole numbers ⺧ {0, 1, 2, 3, 4, 5, . . . } The set of whole numbers includes 0 and the natural numbers. 0, 2, 3, 5, 17 Integers ⺪ {. . . , –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, . . .} The set of integers includes the negatives of the natural numbers and the whole numbers. –17, –5, –3, –2, 0, 2, 3, 5, 17 { ba | a and b are integers and b ⫽ 0} –17= 1 , –5= 1 , –3, –2, 0, 2, 3, 5, 17, Rational numbers ⺡ This means that b is not equal to zero. The set of rational numbers is the set of all numbers that can be expressed as a quotient of two integers, with the denominator not 0. Rational numbers can be expressed as terminating or repeating decimals. Irrational numbers ⺙ Real numbers Rational numbers Integers Irrational numbers Whole numbers Natural numbers Figure P.4 Every real number is either rational or irrational. Examples The set of irrational numbers is the set of all numbers whose decimal representations are neither terminating nor repeating. Irrational numbers cannot be expressed as a quotient of integers. –17 2 =0.4, 5 –2 =–0.6666 3 –5 . . . =–0.6 兹2≠1.414214 –兹3≠–1.73205 p≠3.142 – p ≠–1.571 2 Not all square roots are irrational. For example, 225 = 5 because 52 = 5 # 5 = 25. Thus, 225 is a natural number, a whole number, an integer, and a rational number A 225 = 51 B . The set of real numbers is formed by taking the union of the sets of rational numbers and irrational numbers. Thus, every real number is either rational or irrational, as shown in Figure P.4. Real Numbers The set of real numbers is the set of numbers that are either rational or irrational: 5x ƒ x is rational or x is irrational6. The symbol ⺢ is used to represent the set of real numbers. Thus, ⺢ = 5x ƒ x is rational6 ´ 5x ƒ x is irrational6. EXAMPLE 5 Recognizing Subsets of the Real Numbers Consider the following set of numbers: 3 e - 7, - , 0, 0.6, 25, p, 7.3, 281 f. 4 List the numbers in the set that are a. natural numbers. b. whole numbers. c. integers. d. rational numbers. Solution e. irrational numbers. f. real numbers. a. Natural numbers: The natural numbers are the numbers used for counting. The only natural number in the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F is 281 because 281 = 9. (9 multiplied by itself, or 92, is 81.) 8 Chapter P Prerequisites: Fundamental Concepts of Algebra b. Whole numbers: The whole numbers consist of the natural numbers and 0. The elements of the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F that are whole numbers are 0 and 281. c. Integers: The integers consist of the natural numbers, 0, and the negatives of the natural numbers. The elements of the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F that are integers are 281 , 0, and -7. d. Rational numbers: All numbers in the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F that can be expressed as the quotient of integers are rational numbers. These include - 7 A -7 = -17 B , - 34 , 0 A 0 = 01 B , and 281 A 281 = 91 B . Furthermore, all numbers in the set that are terminating or repeating decimals are also rational numbers. These include 0.6 and 7.3. e. Irrational numbers: The irrational numbers in the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F are 25 A 25 L 2.236 B and p1p L 3.142. Both 25 and p are only approximately equal to 2.236 and 3.14, respectively. In decimal form, 25 and p neither terminate nor have blocks of repeating digits. f. Real numbers: All the numbers in the given set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F are real numbers. Check Point 5 Consider the following set of numbers: e - 9, -1.3, 0, 0.3, p , 29, 210 f. 2 List the numbers in the set that are a. natural numbers. b. whole numbers. d. rational numbers. e. irrational numbers. c. integers. f. real numbers. The Real Number Line The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the origin are positive and numbers to the left of the origin are negative. The real number line is shown in Figure P.5. Negative direction −7 −6 −5 −4 −3 −2 −1 0 1 2 Negative numbers 3 4 5 6 7 Positive direction Positive numbers Figure P.5 The real number line Study Tip Wondering how we located 22 as a precise point on the number line in Figure P.6? We used a right triangle with both legs of length 1. The remaining side measures 22 . 1 Rational numbers 1 兹2 0 Real numbers are graphed on a number line by placing a dot at the correct location for each number. The integers are easiest to locate. In Figure P.6, we’ve graphed six rational numbers and three irrational numbers on a real number line. 1 2 兹2 We’ll have lots more to say about right triangles later in the chapter. Irrational numbers −2 −2 −兹2 ≈ −1.4 − 21 −1 0 0 0.3 = 1 3 1 兹2 ≈1.4 11 4 2 =2 43 3 兹16 =4 4 5 p ≈3.14 Figure P.6 Graphing numbers on a real number line Every real number corresponds to a point on the number line and every point on the number line corresponds to a real number. We say that there is a one-to-one correspondence between all the real numbers and all points on a real number line. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 9 � Use inequality symbols. Ordering the Real Numbers On the real number line, the real numbers increase from left to right. The lesser of two real numbers is the one farther to the left on a number line. The greater of two real numbers is the one farther to the right on a number line. Look at the number line in Figure P.7. The integers - 4 and - 1 are graphed. −5 −4 −3 −2 −1 0 1 2 3 4 5 Figure P.7 Observe that - 4 is to the left of -1 on the number line. This means that -4 is less than -1. −4 is less than −1 because −4 is to the left of −1 on the number line. –4<–1 In Figure P.7, we can also observe that - 1 is to the right of - 4 on the number line. This means that - 1 is greater than -4. –1>–4 −1 is greater than −4 because −1 is to the right of −4 on the number line. The symbols 6 and 7 are called inequality symbols. These symbols always point to the lesser of the two real numbers when the inequality statement is true. −4 is less than −1. −1 is greater than −4. –4<–1 The symbol points to -4, the lesser number. –1>–4 The symbol still points to -4, the lesser number. The symbols 6 and 7 may be combined with an equal sign, as shown in the following table: This inequality is true if either the < part or the = part is true. This inequality is true if either the > part or the = part is true. � Evaluate absolute value. Symbols Meaning Examples Explanation aⱕb a is less than or equal to b. 2ⱕ9 9ⱕ9 Because 2<9 Because 9=9 bⱖa b is greater than or equal to a. 9ⱖ2 2ⱖ2 Because 9>2 Because 2=2 Absolute Value The absolute value of a real number a, denoted by ƒ a ƒ , is the distance from 0 to a on the number line. This distance is always taken to be nonnegative. For example, the real number line in Figure P.8 shows that ƒ -3 ƒ = 3 and ƒ 5 ƒ = 5. 兩−3兩 = 3 兩5兩 = 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 Figure P.8 Absolute value as the distance from 0 The absolute value of -3 is 3 because -3 is 3 units from 0 on the number line. The absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. The absolute value of a positive real number or 0 is the number itself. The absolute value of a negative real number, such as -3, is the number without the negative sign. 10 Chapter P Prerequisites: Fundamental Concepts of Algebra We can define the absolute value of the real number x without referring to a number line. The algebraic definition of the absolute value of x is given as follows: Definition of Absolute Value ƒxƒ = b x -x if x Ú 0 if x 6 0 If x is nonnegative (that is, x Ú 0), the absolute value of x is the number itself. For example, ƒ5ƒ = 5 1 1 ` ` = 3 3 ƒpƒ = p 兩0兩=0. Zero is the only number whose absolute value is 0. If x is a negative number (that is, x 6 0), the absolute value of x is the opposite of x. This makes the absolute value positive. For example, 兩–3兩=–(–3)=3 兩–p兩=–(–p)=p 2 – 1 2 =– a– 1 b = 1 . 3 3 3 This middle step is usually omitted. Evaluating Absolute Value EXAMPLE 6 Rewrite each expression without absolute value bars: a. ƒ 23 - 1 ƒ b. ƒ 2 - p ƒ c. ƒxƒ if x 6 0. x Solution a. Because 23 L 1.7, the number inside the absolute value bars, 23 - 1, is positive. The absolute value of a positive number is the number itself. Thus, ƒ 23 - 1 ƒ = 23 - 1. b. Because p L 3.14, the number inside the absolute value bars, 2 - p, is negative. The absolute value of x when x 6 0 is -x. Thus, ƒ 2 - p ƒ = - 12 - p2 = p - 2. c. If x 6 0, then ƒ x ƒ = - x. Thus, ƒxƒ -x = = - 1. x x Check Point a. ƒ 1 - 22 ƒ 6 Rewrite each expression without absolute value bars: b. ƒ p - 3 ƒ c. ƒxƒ if x 7 0. x Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 11 Listed below are several basic properties of absolute value. Each of these properties can be derived from the definition of absolute value. Discovery Properties of Absolute Value Verify the triangle inequality if a = 4 and b = 5. Verify the triangle inequality if a = 4 and b = - 5. When does equality occur in the triangle inequality and when does inequality occur? Verify your observation with additional number pairs. For all real numbers a and b, 1. ƒ a ƒ Ú 0 2. ƒ -a ƒ = ƒ a ƒ ƒaƒ a 4. ƒ ab ƒ = ƒ a ƒ ƒ b ƒ 5. ` ` = , b Z 0 b ƒbƒ 6. ƒ a + b ƒ … ƒ a ƒ + ƒ b ƒ (called the triangle inequality) � Use absolute value to express distance. 3. a … ƒ a ƒ Distance between Points on a Real Number Line Absolute value is used to find the distance between two points on a real number line. If a and b are any real numbers, the distance between a and b is the absolute value of their difference. For example, the distance between 4 and 10 is 6. Using absolute value, we find this distance in one of two ways: 兩10-4兩=兩6兩=6 or 兩4-10兩=兩–6兩=6. The distance between 4 and 10 on the real number line is 6. Notice that we obtain the same distance regardless of the order in which we subtract. Distance between Two Points on the Real Number Line If a and b are any two points on a real number line, then the distance between a and b is given by ƒ a - b ƒ or ƒ b - a ƒ . Distance between Two Points on a Number Line EXAMPLE 7 Find the distance between -5 and 3 on the real number line. Solution Because the distance between a and b is given by ƒ a - b ƒ , the distance between -5 and 3 is 兩–5-3兩=兩–8兩=8. a = −5 8 −5 −4 −3 −2 −1 0 1 2 3 4 5 Figure P.9 verifies that there are 8 units between -5 and 3 on the real number line. We obtain the same distance if we reverse the order of the subtraction: ƒ 3 - 1-52 ƒ = ƒ 8 ƒ = 8. Figure P.9 The distance between -5 and 3 is 8. Check Point � Identify properties of the real numbers. b=3 7 Find the distance between -4 and 5 on the real number line. Properties of Real Numbers and Algebraic Expressions When you use your calculator to add two real numbers, you can enter them in any order. The fact that two real numbers can be added in any order is called the commutative property of addition. You probably use this property, as well as other 12 Chapter P Prerequisites: Fundamental Concepts of Algebra properties of real numbers listed in Table P.2, without giving it much thought. The properties of the real numbers are especially useful when working with algebraic expressions. For each property listed in Table P.2, a, b, and c represent real numbers, variables, or algebraic expressions. Table P.2 Properties of the Real Numbers Name The Associative Property and the English Language In the English language, phrases can take on different meanings depending on the way the words are associated with commas. Here are three examples. Meaning Commutative Property of Addition Changing order when adding does not affect the sum. a + b = b + a Commutative Property of Multiplication Changing order when multiplying does not affect the product. ab = ba Associative Property of Addition Changing grouping when adding does not affect the sum. 1a + b2 + c = a + 1b + c2 Associative Property of Multiplication Changing grouping when multiplying does not affect the product. 1ab2c = a1bc2 Distributive Property of Multiplication over Addition Multiplication distributes over addition. Examples • 13 + 7 = 7 + 13 • 13x + 7 = 7 + 13x • 22 # 25 = 25 # 22 • x # 6 = 6x • 3 + 18 + x2 = 13 + 82 + x = 11 + x • -213x2 = 1- 2 # 32x = - 6x • 7 A 4+兹3 B =7 ⴢ 4+7 ⴢ 兹3 =28+7兹3 a ⴢ (b+c)=a ⴢ b+a ⴢ c • 5(3x+7)=5 ⴢ 3x+5 ⴢ 7 =15x+35 • 23 + 0 = 23 • 0 + 6x = 6x Identity Property of Addition Zero can be deleted from a sum. a + 0 = a 0 + a = a • What’s the latest dope? What’s the latest, dope? Identity Property of Multiplication • Population of Amsterdam broken down by age and sex Population of Amsterdam, broken down by age and sex One can be deleted from a product. a#1 = a 1#a = a Inverse Property of Addition The sum of a real number and its additive inverse gives 0, the additive identity. a + 1 - a2 = 0 • • • • The product of a nonzero real number and its multiplicative inverse gives 1, the multiplicative identity. 1 a # = 1, a Z 0 a 1# a = 1, a Z 0 a • 7# • Woman, without her man, is nothing. Woman, without her, man is nothing. 1 -a2 + a = 0 Inverse Property of Multiplication • 1#p = p • 13x # 1 = 13x 25 + A - 25 B = 0 -p + p = 0 6x + 1-6x2 = 0 1-4y2 + 4y = 0 1 = 1 7 1 b1x - 32 = 1, x Z 3 • a x - 3 The properties of the real numbers in Table P.2 apply to the operations of addition and multiplication. Subtraction and division are defined in terms of addition and multiplication. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 13 Definitions of Subtraction and Division Let a and b represent real numbers. Subtraction: a - b = a + 1-b2 We call -b the additive inverse or opposite of b. Division: a , b = a # b1 , where b Z 0 We call b1 the multiplicative inverse or reciprocal of b. The quotient of a and b, a , b, can be written in the form ba , where a is the numerator and b the denominator of the fraction. Because subtraction is defined in terms of adding an inverse, the distributive property can be applied to subtraction: a(b-c)=ab-ac (b-c)a=ba-ca. For example, 4(2x-5)=4 ⴢ 2x-4 ⴢ 5=8x-20. 쐅 Simplify algebraic expressions. Simplifying Algebraic Expressions The terms of an algebraic expression are those parts that are separated by addition. For example, consider the algebraic expression 7x - 9y + z - 3, which can be expressed as 7x + 1-9y2 + z + 1- 32. This expression contains four terms, namely 7x, -9y, z, and - 3. The numerical part of a term is called its coefficient. In the term 7x, the 7 is the coefficient. If a term containing one or more variables is written without a coefficient, the coefficient is understood to be 1. Thus, z means 1z. If a term is a constant, its coefficient is that constant. Thus, the coefficient of the constant term -3 is - 3. 7x+(–9y)+z+(–3) Coefficient is 7. Coefficient is −9. Coefficient is 1; z means 1z. Coefficient is −3. The parts of each term that are multiplied are called the factors of the term. The factors of the term 7x are 7 and x. Like terms are terms that have exactly the same variable factors. For example, 3x and 7x are like terms. The distributive property in the form ba + ca = 1b + c2a enables us to add or subtract like terms. For example, Study Tip To combine like terms mentally, add or subtract the coefficients of the terms. Use this result as the coefficient of the terms’ variable factor(s). 3x + 7x = 13 + 72x = 10x 7y 2 - y2 = 7y2 - 1y2 = 17 - 12y2 = 6y2. This process is called combining like terms. An algebraic expression is simplified when parentheses have been removed and like terms have been combined. 14 Chapter P Prerequisites: Fundamental Concepts of Algebra Simplifying an Algebraic Expression EXAMPLE 8 Simplify: 612x2 + 4x2 + 1014x2 + 3x2. Solution 6(2x¤+4x)+10(4x¤+3x) =6 ⴢ 2x¤+6 ⴢ 4x+10 ⴢ 4x¤+10 ⴢ 3x 52x2 and 54x are not like terms. They contain different variable factors, x2 and x, and cannot be combined. =12x¤+24x+40x¤+30x =(12x¤+40x¤)+(24x+30x) =52x¤+54x 8 Check Point Simplify: Use the distributive property to remove the parentheses. Multiply. Group like terms. Combine like terms. 714x2 + 3x2 + 215x2 + x2. Properties of Negatives The distributive property can be extended to cover more than two terms within parentheses. For example, This sign represents subtraction. This sign tells us that the number is negative. –3(4x-2y+6)=–3 ⴢ 4x-(–3) ⴢ 2y-3 ⴢ 6 = - 12x - 1-6y2 - 18 = - 12x + 6y - 18. The voice balloons illustrate that negative signs can appear side by side. They can represent the operation of subtraction or the fact that a real number is negative. Here is a list of properties of negatives and how they are applied to algebraic expressions: Properties of Negatives Let a and b represent real numbers, variables, or algebraic expressions. Property Examples 1. 1 -12a = - a 1 -124xy = - 4xy 2. -1- a2 = a 3. 1- a2b = - ab - 1-6y2 = 6y 1-724xy = - 7 # 4xy = - 28xy 4. a1- b2 = - ab 5x1 -3y2 = - 5x # 3y = - 15xy 5. - 1a + b2 = - a - b -17x + 6y2 = - 7x - 6y 6. - 1a - b2 = - a + b = b - a -13x - 7y2 = - 3x + 7y = 7y - 3x It is not uncommon to see algebraic expressions with parentheses preceded by a negative sign or subtraction. Properties 5 and 6 in the box, - 1a + b2 = - a - b and -1a - b2 = - a + b, are related to this situation. An expression of the form -1a + b2 can be simplified as follows: –(a+b)=–1(a+b)=(–1)a+(–1)b=–a+(–b)=–a-b. Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 15 Do you see a fast way to obtain the simplified expression on the right at the bottom of the previous page? If a negative sign or a subtraction symbol appears outside parentheses, drop the parentheses and change the sign of every term within the parentheses. For example, -13x2 - 7x - 42 = - 3x2 + 7x + 4. Simplifying an Algebraic Expression EXAMPLE 9 Simplify: Solution 8x + 235 - 1x - 324. 8x + 235 - 1x - 324 = 8x + 235 - x + 34 = 8x + 238 - x4 = 8x + 16 - 2x Drop parentheses and change the sign of each term in parentheses: - 1x - 32 = - x + 3. Simplify inside brackets: 5 + 3 = 8. Apply the distributive property: 278 ⴚ x8 ⴝ 2 ⴢ 8 ⴚ 2x ⴝ 16 ⴚ 2x. = 18x - 2x2 + 16 = 18 - 22x + 16 = 6x + 16 Check Point 9 Simplify: Group like terms. Apply the distributive property. Simplify. 6 + 437 - 1x - 224. Using Algebra to Solve Problems in Your Everyday Life Should You Apologize to Your Friend? In this formula, each variable is assigned a number from 1 (low) to 10 (high). How big a deal is the issue? In Geek Logik (Workman Publishing, 2006), humorist Garth Sundem presents formulas covering dating, romance, career, finance, everyday decisions, and health. On the right is a sample of one of his formulas that “takes the guesswork out of life, providing easier living through algebra.” How peeved is your friend? D[Rp(Ra+P)+D(Ra-Rp)]=A How responsible are you for this blunder? How responsible does your friend perceive you to be in this matter? If A is 5 or greater, you should apologize. Exercise Set P.1 Practice Exercises 5. x2 + 3x, for x = 8 6. x2 + 5x, for x = 6 In Exercises 1–16, evaluate each algebraic expression for the given value or values of the variable(s). 7. x2 - 6x + 3, for x = 7 8. x2 - 7x + 4, for x = 8 1. 7 + 5x, for x = 10 2. 8 + 6x, for x = 5 9. 4 + 51x - 723, for x = 9 10. 6 + 51x - 623, for x = 8 3. 6x - y, for x = 3 and y = 8 11. x2 - 31x - y2, for x = 8 and y = 2 4. 8x - y, for x = 3 and y = 4 12. x2 - 41x - y2, for x = 8 and y = 3 16 Chapter P Prerequisites: Fundamental Concepts of Algebra 51x + 22 71x - 32 14. , for x = 10 , for x = 9 2x - 14 2x - 16 2x + 3y 15. , for x = - 2 and y = 4 x + 1 2x + y 16. , for x = - 2 and y = 4 xy - 2x 13. The formula C = 5 1F - 322 9 expresses the relationship between Fahrenheit temperature, F, and Celsius temperature, C. In Exercises 17–18, use the formula to convert the given Fahrenheit temperature to its equivalent temperature on the Celsius scale. 17. 50°F 18. 86°F A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula h = 4 + 60t - 16t2 In Exercises 51–60, rewrite each expression without absolute value bars. 51. ƒ 300 ƒ 52. ƒ - 203 ƒ 55. ƒ 22 - 5 ƒ -3 57. ƒ -3 ƒ 59. 7 -3 ƒ - ƒ -7 7 54. ƒ 7 - p ƒ 56. ƒ 25 - 13 ƒ -7 58. ƒ -7 ƒ 60. 7 - 5 ƒ - ƒ - 137 53. ƒ 12 - p ƒ In Exercises 61–66, evaluate each algebraic expression for x = 2 and y = - 5. 61. ƒ x + y ƒ 62. ƒ x - y ƒ 63. ƒ x ƒ + ƒ y ƒ 64. ƒ x ƒ - ƒ y ƒ 65. y 66. ƒyƒ ƒyƒ ƒxƒ + x y In Exercises 67–74, express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression. describes the ball’s height above the ground, h, in feet, t seconds after it was kicked. Use this formula to solve Exercises 19–20. 67. 2 and 17 68. 4 and 15 69. - 2 and 5 70. - 6 and 8 19. What was the ball’s height 2 seconds after it was kicked? 71. - 19 and - 4 72. - 26 and - 3 20. What was the ball’s height 3 seconds after it was kicked? 73. -3.6 and -1.4 74. -5.4 and -1.2 In Exercises 21–28, find the intersection of the sets. 21. 51, 2, 3, 46 ¨ 52, 4, 56 23. 5s, e, t6 ¨ 5t, e, s6 25. 51, 3, 5, 76 ¨ 52, 4, 6, 8, 106 26. 50, 1, 3, 56 ¨ 5 -5, -3, - 16 27. 5a, b, c, d6 ¨ ⭋ 22. 51, 3, 76 ¨ 52, 3, 86 24. 5r, e, a, l6 ¨ 5l, e, a, r6 28. 5w, y, z6 ¨ ⭋ In Exercises 29–34, find the union of the sets. 29. 51, 2, 3, 46 ´ 52, 4, 56 30. 51, 3, 7, 86 ´ 52, 3, 86 33. 5a, e, i, o, u6 ´ ⭋ 34. 5e, m, p, t, y6 ´ ⭋ 31. 51, 3, 5, 76 ´ 52, 4, 6, 8, 106 32. 50, 1, 3, 56 ´ 52, 4, 66 In Exercises 35–38, list all numbers from the given set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers. 35. 36. 37. 38. E - 9, - 54 , 0, 0.25, 23 , 9.2, 2100 F E - 7, -0.6, 0, 249, 250 F E - 11, - 65 , 0, 0.75, 25 , p, 264 F E - 5, -0.3, 0, 22, 24 F 39. Give an example of a whole number that is not a natural number. 40. Give an example of a rational number that is not an integer. 41. Give an example of a number that is an integer, a whole number, and a natural number. 42. Give an example of a number that is a rational number, an integer, and a real number. Determine whether each statement in Exercises 43–50 is true or false. 43. - 13 … - 2 44. - 6 7 2 45. 4 Ú - 7 46. - 13 6 - 5 47. - p Ú - p 48. - 3 7 - 13 49. 0 Ú - 6 50. 0 Ú - 13 In Exercises 75–84, state the name of the property illustrated. 75. 6 + 1 -42 = 1-42 + 6 76. 11 # 17 + 42 = 11 # 7 + 11 # 4 77. 6 + 12 + 72 = 16 + 22 + 7 78. 6 # 12 # 32 = 6 # 13 # 22 79. 12 + 32 + 14 + 52 = 14 + 52 + 12 + 32 80. 7 # 111 # 82 = 111 # 82 # 7 81. 21- 8 + 62 = - 16 + 12 82. -813 + 112 = - 24 + 1-882 83. 1 1x + 32 = 1, x Z - 3 1x + 32 84. 1x + 42 + 3- 1x + 424 = 0 In Exercises 85–96, simplify each algebraic expression. 85. 513x + 42 - 4 86. 215x + 42 - 3 87. 513x - 22 + 12x 88. 215x - 12 + 14x 89. 713y - 52 + 214y + 32 90. 412y - 62 + 315y + 102 93. 7 - 433 - 14y - 524 94. 6 - 538 - 12y - 424 91. 513y - 22 - 17y + 22 92. 415y - 32 - 16y + 32 95. 18x + 4 - 361x - 22 + 54 2 2 96. 14x2 + 5 - 371x2 - 22 + 44 In Exercises 97–102, write each algebraic expression without parentheses. 97. -1- 14x2 99. -12x - 3y - 62 101. 1313x2 + 314y2 + 1- 4y24 102. 1212y2 + 31 -7x2 + 7x4 98. - 1- 17y2 100. -15x - 13y - 12 Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers Practice Plus In Exercises 103–110, insert either 6, 7 , or = in the shaded area to make a true statement. 103. ƒ - 6 ƒ ƒ -3 ƒ 3 105. ` ` 5 ƒ -0.6 ƒ 5 106. ` ` 2 14 # 15 15 14 3 30 107. 40 4 109. 104. ƒ - 20 ƒ 8 8 , 13 13 ƒ - 50 ƒ ƒ - 2.5 ƒ 17 # 18 108. 18 17 ƒ -1 ƒ 5 50 60 6 4 4 , 17 17 110. ƒ - 2 ƒ limits of these ranges are fractions of the maximum heart rate, 220 - a. Exercises 129–130 are based on the information in the graph. Target Heart Rate Ranges for Exercise Goals Exercise Goal Boost performance as a competitive athlete Improve cardiovascular conditioning Lose weight Improve overall health and reduce risk of heart attack W In Exercises 111–120, use the order of operations to simplify each expression. 111. 82 - 16 , 2 2 # 4 - 3 113. 112. 102 - 100 , 52 # 2 - 3 5 # 2 - 32 332 - 1-2242 114. 10 , 2 + 3 # 4 112 - 3 # 222 q 118. 5 - 8 15 - 62 - 2 ƒ 3 - 7 ƒ 2 119. 89 - 3 # 52 120. Lower limit of range Upper limit of range 61 -42 - 51-32 Lower limit of range Upper limit of range 122. A number decreased by the difference between eight and the number 7 (220-a) 10 4 H= (220-a). 5 H= 1 (220-a) 2 3 H= (220-a). 5 H= a. What is the lower limit of the heart range, in beats per minute, for a 30-year-old with this exercise goal? 123. Six times the product of negative five and a number 124. Ten times the product of negative four and a number 126. The difference between the product of six and a number and negative two times the number b. What is the upper limit of the heart range, in beats per minute, for a 30-year-old with this exercise goal? The bar graph shows the average cost of tuition and fees at private four-year colleges in the United States. Average Cost of Tuition and Fees at Private Four-Year United States Colleges 127. The difference between eight times a number and six more than three times the number $23,000 128. Eight decreased by three times the sum of a number and six 22,218 $22,000 21,235 The maximum heart rate, in beats per minute, that you should achieve during exercise is 220 minus your age: 220-a. Tuition and Fees $21,000 Application Exercises The bar graph at the top of the next column shows the target heart rate ranges for four types of exercise goals. The lower and upper 19,710 $20,000 $19,000 20,082 18,273 $18,000 17,272 $17,000 $16,000 This algebraic expression gives maximum heart rate in terms of age, a. 1 130. If your exercise goal is to improve overall health, the graph shows the following range for target heart rate, H, in beats per minute: 121. A number decreased by the sum of the number and four 125. The difference between the product of five and a number and twice the number Ï b. What is the upper limit of the heart range, in beats per minute, for a 20-year-old with this exercise goal? 12 , 3 # 5 ƒ 2 2 + 32 ƒ In Exercises 121–128, write each English phrase as an algebraic expression. Then simplify the expression. Let x represent the number. R a. What is the lower limit of the heart range, in beats per minute, for a 20-year-old with this exercise goal? 9 - 10 7 + 3 - 62 Î 129. If your exercise goal is to improve cardiovascular conditioning, the graph shows the following range for target heart rate, H, in beats per minute: 116. 8 - 33- 215 - 72 - 514 - 224 21- 22 - 41-32 E Fraction of Maximum Heart Rate, 220 − a 115. 8 - 33-212 - 52 - 418 - 624 117. 17 16,233 15,518 $15,000 2000 2001 2002 2003 2004 2005 2006 2007 Ending Year in the School Year Source: The College Board 18 Chapter P Prerequisites: Fundamental Concepts of Algebra The formula T = 15,395 + 988x - 2x 2 models the average cost of tuition and fees, T, at private U.S. colleges for the school year ending x years after 2000. Use this information to solve Exercises 131–132. 131. a. Use the formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2007. b. By how much does the formula underestimate or overestimate the actual cost shown by the graph at the bottom of the previous page for the school year ending in 2007? c. Use the formula to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2010. 132. a. Use the formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2006. b. By how much does the formula underestimate or overestimate the actual cost shown by the graph at the bottom of the previous page for the school year ending in 2006? c. Use the formula to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2012. 133. You had $10,000 to invest. You put x dollars in a safe, government-insured certificate of deposit paying 5% per year. You invested the remainder of the money in noninsured corporate bonds paying 12% per year. Your total interest earned at the end of the year is given by the algebraic expression 0.05x + 0.12110,000 - x2. a. Simplify the algebraic expression. b. Use each form of the algebraic expression to determine your total interest earned at the end of the year if you invested $6000 in the safe, government-insured certificate of deposit. 134. It takes you 50 minutes to get to campus.You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression 136. If n is a natural number, what does bn mean? Give an example with your explanation. 137. What does it mean when we say that a formula models realworld phenomena? 138. What is the intersection of sets A and B? 139. What is the union of sets A and B? 140. How do the whole numbers differ from the natural numbers? 141. Can a real number be both rational and irrational? Explain your answer. 142. If you are given two real numbers, explain how to determine which is the lesser. 143. Think of a situation where you either apologized or did not apologize for a blunder you committed. Use the formula in the essay on page 15 to determine whether or not you should have apologized. How accurately does the formula model what you actually did? 144. Read Geek Logik by Garth Sundem (Workman Publishing, 2006). Would you recommend the book to college algebra students? Were you amused by the humor? Which formulas did you find most useful? Apply at least one of the formulas by plugging your life data into the equation, using the order of operations, and coming up with an answer to one of life’s dilemmas. Critical Thinking Exercises Make Sense? In Exercises 145–148, determine whether each statement makes sense or does not make sense, and explain your reasoning. 145. My mathematical model describes the data for tuition and fees at public four-year colleges for the past ten years extremely well, so it will serve as an accurate prediction for the cost of public colleges in 2050. 146. A model that describes the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000 cannot be used to estimate the cost of private education for the school year ending in 2000. 147. The humor in this cartoon is based on the fact that the football will never be hiked. 0.06t + 0.5150 - t2. a. Simplify the algebraic expression. b. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop. Writing in Mathematics Writing about mathematics will help you learn mathematics. For all writing exercises in this book, use complete sentences to respond to the question. Some writing exercises can be answered in a sentence; others require a paragraph or two. You can decide how much you need to write as long as your writing clearly and directly answers the question in the exercise. Standard references such as a dictionary and a thesaurus should be helpful. 135. What is an algebraic expression? Give an example with your explanation. FOXTROT © 2000 Bill Amend. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved. 148. Just as the commutative properties change groupings, the associative properties change order. In Exercises 149–156, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 149. Every rational number is an integer. 150. Some whole numbers are not integers. 151. Some rational numbers are not positive. Section P.2 Exponents and Scientific Notation 19 b. b5 # b5 = 1b # b # b # b # b21b # b # b # b # b2 = b? 152. Irrational numbers cannot be negative. 153. The term x has no coefficient. c. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base? 154. 5 + 31x - 42 = 81x - 42 = 8x - 32 155. -x - x = - x + 1-x2 = 0 156. x - 0.021x + 2002 = 0.98x - 4 161. In parts (a) and (b), complete each statement. In Exercises 157–159, insert either 6 or 7 in the shaded area between the numbers to make the statement true. 157. 22 3.14 159. 2 p 2 Preview Exercises 160. In parts (a) and (b), complete each statement. a. b4 # b3 = 1b # b # b # b21b # b # b2 = b? P.2 b. b#b#b#b#b#b#b#b b8 = b? = 2 b#b b c. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base? Exercises 160–162 will help you prepare for the material covered in the next section. Section b#b#b#b#b#b#b b7 = b? = 3 b#b#b b - 3.5 158. - p 1.5 a. 162. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2? Exponents and Scientific Notation Objectives � Use properties of exponents. � Simplify exponential expressions. � Use scientific notation. � Use properties of exponents. L istening to the radio on the way to campus, you hear politicians discussing the problem of the national debt, which exceeds $9 trillion. They state that it’s more than the gross domestic product of China, the world’s secondrichest nation, and four times greater than the combined net worth of America’s 691 billionaires. They make it seem like the national debt is a real problem, but later you realize that you don’t really know what a number like 9 trillion means. If the national debt were evenly divided among all citizens of the country, how much would every man, woman, and child have to pay? Is economic doomsday about to arrive? In this section, you will learn to use exponents to provide a way of putting large and small numbers in perspective. Using this skill, we will explore the meaning of the national debt. Properties of Exponents The major properties of exponents are summarized in the box that follows and continues on the next page. Properties of Exponents Property Study Tip The Negative-Exponent Rule When a negative integer appears as an exponent, switch the position of the base (from numerator to denominator or from denominator to numerator) and make the exponent positive. If b is any real number other than 0 and n is a natural number, then 1 b-n = n . b Examples • 5-3 = • 1 1 = 3 125 5 1 1 = = 42 = 16 -2 1 4 42 Section P.2 Exponents and Scientific Notation 19 b. b5 # b5 = 1b # b # b # b # b21b # b # b # b # b2 = b? 152. Irrational numbers cannot be negative. 153. The term x has no coefficient. c. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base? 154. 5 + 31x - 42 = 81x - 42 = 8x - 32 155. -x - x = - x + 1-x2 = 0 156. x - 0.021x + 2002 = 0.98x - 4 161. In parts (a) and (b), complete each statement. In Exercises 157–159, insert either 6 or 7 in the shaded area between the numbers to make the statement true. 157. 22 3.14 159. 2 p 2 Preview Exercises 160. In parts (a) and (b), complete each statement. a. b4 # b3 = 1b # b # b # b21b # b # b2 = b? P.2 b. b#b#b#b#b#b#b#b b8 = b? = 2 b#b b c. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base? Exercises 160–162 will help you prepare for the material covered in the next section. Section b#b#b#b#b#b#b b7 = b? = 3 b#b#b b - 3.5 158. - p 1.5 a. 162. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2? Exponents and Scientific Notation Objectives � Use properties of exponents. � Simplify exponential expressions. � Use scientific notation. � Use properties of exponents. L istening to the radio on the way to campus, you hear politicians discussing the problem of the national debt, which exceeds $9 trillion. They state that it’s more than the gross domestic product of China, the world’s secondrichest nation, and four times greater than the combined net worth of America’s 691 billionaires. They make it seem like the national debt is a real problem, but later you realize that you don’t really know what a number like 9 trillion means. If the national debt were evenly divided among all citizens of the country, how much would every man, woman, and child have to pay? Is economic doomsday about to arrive? In this section, you will learn to use exponents to provide a way of putting large and small numbers in perspective. Using this skill, we will explore the meaning of the national debt. Properties of Exponents The major properties of exponents are summarized in the box that follows and continues on the next page. Properties of Exponents Property Study Tip The Negative-Exponent Rule When a negative integer appears as an exponent, switch the position of the base (from numerator to denominator or from denominator to numerator) and make the exponent positive. If b is any real number other than 0 and n is a natural number, then 1 b-n = n . b Examples • 5-3 = • 1 1 = 3 125 5 1 1 = = 42 = 16 -2 1 4 42 20 Chapter P Prerequisites: Fundamental Concepts of Algebra The Zero-Exponent Rule If b is any real number other than 0, • 70 = 1 • 1-520 = 1 b0 = 1. • –50=–1 Only 5 is raised to the zero power. The Product Rule If b is a real number or algebraic expression, and m and n are integers, b m #b n m+n = b • 22 # 23 = 22 + 3 = 25 = 32 • x-3 # x7 = x-3 + 7 = x4 . When multiplying exponential expressions with the same base, add the exponents. Use this sum as the exponent of the common base. The Power Rule If b is a real number or algebraic expression, and m and n are integers, 1bm2 = bmn. n # 3 • 1222 = 22 3 = 26 = 64 1 # 4 • 1x-32 = x-3 4 = x-12 = 12 x When an exponential expression is raised to a power, multiply the exponents. Place the product of the exponents on the base and remove the parentheses. Study Tip The Quotient Rule 45 represent different 43 4 numbers: If b is a nonzero real number or algebraic expression, and m and n are integers, • 28 = 28 - 4 = 24 = 16 24 bm = bm - n. bn • 1 x3 = x3 - 7 = x-4 = 4 7 x x 43 5 and 4 3 45 = 43 - 5 = 4-2 = 1 1 = 16 42 45 = 45 - 3 = 42 = 16. 43 When dividing exponential expressions with the same nonzero base, subtract the exponent in the denominator from the exponent in the numerator. Use this difference as the exponent of the common base. Products Raised to Powers If a and b are real numbers or algebraic expressions, and n is an integer, 1ab2n = anbn. • 1-2y24 = 1 -224 y4 = 16y4 • 1-2xy23 = 1 -223 x3 y3 = - 8x3 y3 When a product is raised to a power, raise each factor to that power. Quotients Raised to Powers If a and b are real numbers, b Z 0, or algebraic expressions, and n is an integer, a n an a b = n. b b 24 16 2 4 • a b = 4 = 5 625 5 1- 323 27 3 3 = - 3 • a- b = x x3 x When a quotient is raised to a power, raise the numerator to that power and divide by the denominator to that power. Section P.2 Exponents and Scientific Notation 21 � Simplify exponential expressions. Simplifying Exponential Expressions Properties of exponents are used to simplify exponential expressions. An exponential expression is simplified when • No parentheses appear. • No powers are raised to powers. • Each base occurs only once. • No negative or zero exponents appear. Simplifying Exponential Expressions Example 1. If necessary, remove parentheses by using a n an 1ab2n = anbn or a b = n . b b 1xy23 = x3y3 2. If necessary, simplify powers to powers by using 1bm2 = bmn. # 3 1x42 = x4 3 = x12 n 3. If necessary, be sure that each base appears only once by using bm bm # bn = bm + n or n = bm - n. b 4. If necessary, rewrite exponential expressions with zero powers as 1 1b0 = 12. Furthermore, write the answer with positive exponents by using 1 1 b-n = n or -n = bn. b b x4 # x3 = x4 + 3 = x7 1 x5 = x5 - 8 = x -3 = 3 8 x x The following example shows how to simplify exponential expressions. Throughout the example, assume that no variable in a denominator is equal to zero. EXAMPLE 1 Simplifying Exponential Expressions Simplify: a. 1- 3x y 2 4 5 3 b. 1-7xy 21 - 2x y 2 4 5 6 c. - 35x2 y4 5x6y -8 -3 4x2 d. ¢ ≤ . y Solution a. 1-3x4 y52 = 1-3231x42 1y52 3 3 3 # # = 1-323 x4 3 y5 3 = - 27x12y15 Raise each factor inside the parentheses to the third power. Multiply the exponents when raising powers to powers. 1-323 = 1-321 -321 -32 = - 27 b. 1-7xy421 - 2x5y 62 = 1 -721 -22xx5 y4y6 Group factors with the same base. = 14x1 + 5y4 + 6 When multiplying expressions with the same base, add the exponents. = 14x6y10 Simplify. 22 Chapter P Prerequisites: Fundamental Concepts of Algebra -35x2y4 c. = a 6 -8 5x y y4 -35 x2 b ¢ 6 ≤ ¢ -8 ≤ 5 x y = - 7x2 - 6y4 - 1-82 4x2 ≤ y When dividing expressions with the same base, subtract the exponents. = - 7x -4 y12 Simplify. Notice that 4 - 1-82 = 4 + 8 = 12. -7y12 Write as a fraction and move the base with the negative exponent, x-4, to the other side of the fraction bar and make the negative exponent positive. = d. ¢ Group factors with the same base. x4 = 14x22 = 4 -31x22 -3 = = = Check Point a. 12x3y62 4 -3 Raise the numerator and the denominator to the -3 power. y -3 -3 y -3 4 -3 x -6 y -3 Multiply the exponents when raising a power to a power: 1x22-3 = x21-32 = x-6. y3 Move each base with a negative exponent to the other side of the fraction bar and make each negative exponent positive. 3 6 4 x y3 43 = 4 # 4 # 4 = 64 64x6 1 Raise each factor in the numerator to the - 3 power. Simplify: b. 1 - 6x2y5213xy32 c. 100x 12 y2 20x16 y -4 d. ¢ 5x -2 ≤ . y4 Study Tip Try to avoid the following common errors that can occur when simplifying exponential expressions. Correct #b 32 # 34 b 3 4 = b Incorrect Description of Error #b = b 32 # 34 = 96 7 b = 36 3 4 12 The exponents should be added, not multiplied. The common base should be retained, not multiplied. 516 = 512 54 516 = 54 54 The exponents should be subtracted, not divided. 14a23 = 64a3 14a23 = 4a3 Both factors should be cubed. b-n = 1 bn 1a + b2-1 = b-n = 1 a + b 1 bn 1a + b2-1 = Only the exponent should change sign. 1 1 + a b The exponent applies to the entire expression a + b. Section P.2 Exponents and Scientific Notation 23 � Use scientific notation. Table P.3 Names of Large Numbers 102 hundred 10 3 thousand 10 6 million Scientific Notation As of December 2007, the national debt of the United States was about $9.2 trillion. This is the amount of money the government has had to borrow over the years, mostly by selling bonds, because it has spent more than it has collected in taxes. A stack of $1 bills equaling the national debt would rise to twice the distance from Earth to the moon. Because a trillion is 1012 (see Table P.3), the national debt can be expressed as 9.2 * 1012. The number 9.2 * 1012 is written in a form called scientific notation. 109 billion 1012 trillion Scientific Notation quadrillion A number is written in scientific notation when it is expressed in the form 10 15 1018 quintillion 1021 sextillion 1024 septillion 1027 octillion 10 30 nonillion 10 100 googol 10googol googolplex a * 10n, where the absolute value of a is greater than or equal to 1 and less than 10 11 … ƒ a ƒ 6 102, and n is an integer. It is customary to use the multiplication symbol, * , rather than a dot, when writing a number in scientific notation. Converting from Scientific to Decimal Notation Here are two examples of numbers in scientific notation: 6.4 * 105 means 640,000. 2.17 * 10-3 means 0.00217. Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small? We can use n, the exponent on the 10 in a * 10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left ƒ n ƒ places. Converting from Scientific to Decimal Notation EXAMPLE 2 Write each number in decimal notation: a. 6.2 * 107 b. -6.2 * 107 c. 2.019 * 10 -3 d. - 2.019 * 10-3. Solution In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left. a. 6.2*10‡=62,000,000 n=7 n=7 Move the decimal point 7 places to the right. c. 2.019*10–‹=0.002019 n = −3 Check Point b. –6.2*10‡=–62,000,000 d. –2.019*10–‹=–0.002019 Move the decimal point 兩−3兩 places, or 3 places, to the left. 2 a. -2.6 * 10 9 Move the decimal point 7 places to the right. n = −3 Move the decimal point 兩−3兩 places, or 3 places, to the left. Write each number in decimal notation: b. 3.017 * 10-6. 24 Chapter P Prerequisites: Fundamental Concepts of Algebra Converting from Decimal to Scientific Notation To convert from decimal notation to scientific notation, we reverse the procedure of Example 2. Converting from Decimal to Scientific Notation Write the number in the form a * 10n. • Determine a, the numerical factor. Move the decimal point in the given number to obtain a number whose absolute value is between 1 and 10, including 1. • Determine n, the exponent on 10n. The absolute value of n is the number of places the decimal point was moved. The exponent n is positive if the decimal point was moved to the left, negative if the decimal point was moved to the right, and 0 if the decimal point was not moved. Converting from Decimal Notation to Scientific Notation EXAMPLE 3 Write each number in scientific notation: b. -34,970,000,000,000 d. -0.0000000000802. a. 34,970,000,000,000 c. 0.0000000000802 Solution Technology a. 34,970,000,000,000=3.497*10⁄‹ You can use your calculator’s 冷EE 冷 (enter exponent) or 冷EXP 冷 key to convert from decimal to scientific notation. Here is how it’s done for 0.0000000000802. Many Scientific Calculators Keystrokes Move the decimal point to get a number whose absolute value is between 1 and 10. The decimal point was moved 13 places to the left, so n = 13. b. –34,970,000,000,000=–3.497*10⁄‹ c. 0.0000000000802=8.02*10–⁄⁄ .0000000000802 冷EE 冷 冷 = Study Tip If the absolute value of a number is greater than 10, it will have a positive exponent in scientific notation. If the absolute value of a number is less than 1, it will have a negative exponent in scientific notation. 冷 Display 8.02 - 11 Move the decimal point to get a number whose absolute value is between 1 and 10. The decimal point was moved 11 places to the right, so n = −11. Many Graphing Calculators d. –0.0000000000802=–8.02*10–⁄⁄ Use the mode setting for scientific notation. Check Point 3 Write each number in scientific notation: Keystrokes .0000000000802 冷ENTER 冷 a. 5,210,000,000 b. -0.00000006893. Display 8.02E - 11 EXAMPLE 4 Expressing the U.S. Population in Scientific Notation As of December 2007, the population of the United States was approximately 303 million. Express the population in scientific notation. Solution Because a million is 106, the 2007 population can be expressed as 303*10fl. This factor is not between 1 and 10, so the number is not in scientific notation. Section P.2 Exponents and Scientific Notation 25 The voice balloon indicates that we need to convert 303 to scientific notation. 303*10fl=(3.03*10¤)*10fl=3.03*10¤± fl=3.03*10° 303 = 3.03 ×102 In scientific notation, the population is 3.03 * 108. Check Point 4 Express 410 * 107 in scientific notation. Computations with Scientific Notation Properties of exponents are used to perform computations with numbers that are expressed in scientific notation. EXAMPLE 5 Computations with Scientific Notation Perform the indicated computations, writing the answers in scientific notation: a. 16.1 * 105214 * 10-92 b. 1.8 * 104 . 3 * 10-2 Solution a. 16.1 * 105214 * 10-92 Technology 16.1 * 105214 * 10-92 On a Calculator: = 16.1 * 42 * 1105 * 10-92 = 24.4 * 105 + 1-92 Many Scientific Calculators 6.1 冷EE 冷 5 冷 * 冷 4 冷EE 冷 9 冷 +/ - 冷 冷 = 冷 Convert 24.4 to scientific notation: 24.4 = 2.44 * 101. 101 * 10-4 = 101 + 1-42 = 10-3 = 2.44 * 10-3 Many Graphing Calculators 6.1 冷EE 冷 5 冷 * 冷 4 冷EE 冷 冷1-2 冷 9 冷ENTER 冷 2.44E - 3 Simplify. = 12.44 * 1012 * 10-4 2.44 - 03 Display (in scientific notation mode) Add the exponents on 10 and multiply the other parts. = 24.4 * 10-4 Display b. 1.8 * 104 1.8 104 = a b * ¢ -2 ≤ -2 3 3 * 10 10 = 0.6 * 104 - 1-22 = 0.6 * 106 = 16 * 10-12 * 106 = 6 * 105 Check Point 5 Regroup factors. Regroup factors. Subtract the exponents on 10 and divide the other parts. Simplify: 4 - 1 - 22 = 4 + 2 = 6. Convert 0.6 to scientific notation: 0.6 = 6 * 10-1. 10-1 * 106 = 10-1 + 6 = 105 Perform the indicated computations, writing the answers in scientific notation: a. 17.1 * 105215 * 10-72 b. 1.2 * 106 . 3 * 10-3 26 Chapter P Prerequisites: Fundamental Concepts of Algebra Applications: Putting Numbers in Perspective Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $9.2 trillion as of December 2007. The graph in Figure P.10 shows the national debt increasing over time. National Debt (trillions of dollars) The National Debt Figure P.10 Source: Office of Management and Budget 10 9 8 7 9.2 6 5 4 3 2 1 0 4.9 5.2 5.4 5.5 5.6 5.7 5.8 6.2 6.8 7.4 7.9 8.5 3.2 1.8 0.9 1980 1985 1990 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 Year Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 9.2 trillion. EXAMPLE 6 The National Debt As of December 2007, the national debt was $9.2 trillion, or 9.2 * 1012 dollars. At that time, the U.S. population was approximately 303,000,000 (303 million), or 3.03 * 108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay? Solution The amount each citizen must pay is the total debt, 9.2 * 1012 dollars, divided by the number of citizens, 3.03 * 108. 9.2 * 1012 9.2 1012 = a b * ¢ ≤ 3.03 3.03 * 108 108 L 3.04 * 1012 - 8 = 3.04 * 104 = 30,400 Every U.S. citizen would have to pay approximately $30,400 to the federal government to pay off the national debt. Check Point 6 Pell Grants help low-income undergraduate students pay for college. In 2006, the federal cost of this program was $13 billion 1$13 * 1092 and there were 5.1 million 15.1 * 1062 grant recipients. How much, to the nearest hundred dollars, was the average grant? An Application: Black Holes in Space The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black. Section P.2 Exponents and Scientific Notation 27 A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example. EXAMPLE 7 An Application of Scientific Notation Use the Schwarzchild formula Rs = where 2GM c2 Rs = Radius of the star, in meters, that would cause it to become a black hole M = Mass of the star, in kilograms G = A constant, called the gravitational constant m3 = 6.7 * 10-11 kg # s2 c = Speed of light = 3 * 108 meters per second to determine to what length the radius of the sun must be reduced for it to become a black hole. The sun’s mass is approximately 2 * 1030 kilograms. Solution Rs = = = = = = 2GM c2 Use the given model. 2 * 6.7 * 10-11 * 2 * 1030 13 * 1082 2 G = 6.7 * 10-11, M = 2 * 1030, and c = 3 * 108. 12 * 6.7 * 22 * 110-11 * 10302 13 * 1082 2 26.8 * 10-11 + 30 32 * 11082 2 26.8 * 1019 9 * 1016 Rearrange factors in the numerator. Add exponents in the numerator. Raise each factor in the denominator to the power. Multiply powers to powers: 11082 = 108 2 = 1016. 2 26.8 * 1019 - 16 9 L 2.978 * 103 Substitute the given values: # When dividing expressions with the same base, subtract the exponents. Simplify. = 2978 Although the sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the sun’s radius were reduced to approximately 2978 meters, that is, about 1 235,000 its present size, it would become a black hole. 7 The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by Check Point S = 11.76 * 1052311.44 * 10-22 - r24. Find the speed of blood at the central axis of this artery. 28 Chapter P Prerequisites: Fundamental Concepts of Algebra Exercise Set P.2 Practice Exercises Evaluate each exponential expression in Exercises 1–22. 1. 52 # 2 2. 62 # 2 3. 1 - 22 4. 1 -22 6 5. -2 4 6 6. -2 4 7. 1 - 320 8. 1 - 920 9. -30 5x3 ≤ y 61. ¢ -15a4 b2 ≤ 5a10 b-3 63. ¢ 3a -5 b2 ≤ 12a3 b-4 3 0 -3 60. ¢ 3x4 ≤ y 62. ¢ - 30a14 b8 ≤ 10a17 b-2 64. ¢ 4a -5 b3 3 -5 12a b ≤ 3 0 10. - 90 11. 4 -3 12. 2 -6 13. 2 2 # 2 3 14. 33 # 32 15. 12 22 16. 1332 3 In Exercises 65–76, write each number in decimal notation without the use of exponents. 2 8 17. -2 59. ¢ 8 2 24 18. 3 34 19. 3 -3 # 3 20. 2 -3 # 2 23 21. 7 2 34 22. 7 3 Simplify each exponential expression in Exercises 23–64. -2 23. x y 24. xy -3 7 0 0 5 65. 3.8 * 102 66. 9.2 * 102 67. 6 * 10-4 68. 7 * 10-5 69. -7.16 * 106 70. - 8.17 * 106 71. 7.9 * 10-1 72. 6.8 * 10-1 73. -4.15 * 10-3 74. -3.14 * 10-3 75. -6.00001 * 1010 76. - 7.00001 * 1010 In Exercises 77–86, write each number in scientific notation. 77. 32,000 78. 64,000 25. x y 26. x y 79. 638,000,000,000,000,000 80. 579,000,000,000,000,000 27. x3 # x7 28. x11 # x5 81. - 5716 82. - 3829 29. x -5 # x10 30. x -6 # x12 83. 0.0027 84. 0.0083 31. 1x 2 32. 1x 2 85. -0.00000000504 86. -0.00000000405 x14 35. 7 x x30 36. 10 x x14 37. -7 x x30 38. -10 x 3 7 11 5 33. 1x 2 34. 1x 2 -5 3 -6 4 39. 18x 2 4 2 3 6 42. a - b y 3 43. 1 - 3x2y52 44. 1 -3x4 y62 47. 1- 9x3y21 - 2x6y42 48. 1 -5x4y21 -6x7 y112 2 45. 13x4212x 72 49. 8x20 2x4 46. 111x5219x122 50. 25a b - 5a2b3 55. 14x32 57. 10x6 52. 35a b - 7a7b3 54. -2 24x3 y5 7 -9 32x y 92. 15.1 * 10-8213 * 10-42 93. 14.3 * 108216.2 * 1042 20b 10b20 56. 110x22 58. 95. 97. 99. 101. 10 14b 7b14 91. 16.1 * 10 -8212 * 10-42 20x24 14 6 7 53. 88. 12 * 104214.1 * 1032 3 13 4 51. 87. 13 * 104212.1 * 1032 89. 11.6 * 1015214 * 10-112 40. 16x 2 3 2 4 41. a - b x In Exercises 87–106, perform the indicated computations. Write the answers in scientific notation. If necessary, round the decimal factor in your scientific notation answer to two decimal places. -3 10x 4 y9 30x12y -3 8.4 * 108 4 * 105 3.6 * 104 9 * 10-2 4.8 * 10-2 2.4 * 10 6 2.4 * 10-2 4.8 * 10 -6 90. 11.4 * 1015213 * 10-112 94. 18.2 * 108214.6 * 1042 96. 98. 100. 102. 6.9 * 108 3 * 105 1.2 * 104 2 * 10-2 7.5 * 10-2 2.5 * 106 1.5 * 10 -2 3 * 10-6 103. 480,000,000,000 0.00012 104. 282,000,000,000 0.00141 105. 0.00072 * 0.003 0.00024 106. 66,000 * 0.001 0.003 * 0.002 Section P.2 Exponents and Scientific Notation 29 Practice Plus In Exercises 107–114, simplify each exponential expression. Assume that variables represent nonzero real numbers. -2 -3 1xy -22 1x -2 y2 107. 108. 3 -3 1x2y -12 1x -2 y2 x -3 y -4 z-5 ≤ -2 112. ¢ -4 3 -2 12x -3 y -52 ≤ Tickets Sold 1600 -3 3 0 2 12 -1 x -3 y -12 12x -6 y42 19x3y -32 -2 114. x -4 y -5 z-6 United States Film Admissions and Admission Charges -4 12 x y 2 12x y 2 116x y 2 -1 -2 -1 -2 113. x 4 y 5 z6 -2 0 12x y 2 -4 -6 2 Application Exercises The bar graph shows the total amount Americans paid in federal taxes, in trillions of dollars, and the U.S. population, in millions, from 2003 through 2006. Exercises 115–116 are based on the numbers displayed by the graph. Annual Admissions (billions of tickets sold) 111. ¢ x 3 y 4 z5 110. 13x -4 yz-7213x2-3 1400 1200 1000 Price per Ticket 1420 1400 1260 1190 $8 1450 6.40 6.60 4.20 $7 $6 5.40 $5 4.40 800 $4 600 $3 400 $2 200 $1 1990 1995 2000 Year 2005 Average Price per Ticket 109. 12x -3 yz-6212x2-5 In the dramatic arts, ours is the era of the movies. As individuals and as a nation, we’ve grown up with them. Our images of love, war, family, country—even of things that terrify us—owe much to what we’ve seen on screen. The bar graph quantifies our love for movies by showing the number of tickets sold, in billions, and the average price per ticket for five selected years. Exercises 117–118 are based on the numbers displayed by the graph. 2006 Source: National Association of Theater Owners Federal Taxes and the United States Population Federal Taxes Collected Population Federal Taxes Collected (trillions of dollars) $2.00 295 292 $2.50 1.95 2.27 298 2.52 300 2.02 300 250 $1.50 200 $1.00 150 $0.50 100 2003 2005 2004 Population (millions) 350 $3.00 2006 Year Sources: Internal Revenue Service and U.S. Census Bureau 115. a. In 2006, the United States government collected $2.52 trillion in taxes. Express this number in scientific notation. b. In 2006, the population of the United States was approximately 300 million. Express this number in scientific notation. c. Use your scientific notation answers from parts (a) and (b) to answer this question: If the total 2006 tax collections were evenly divided among all Americans, how much would each citizen pay? Express the answer in decimal notation. 116. a. In 2005, the United States government collected $2.27 trillion in taxes. Express this number in scientific notation. b. In 2005, the population of the United States was approximately 298 million. Express this number in scientific notation. c. Use your scientific notation answers from parts (a) and (b) to answer this question: If the total 2005 tax collections were evenly divided among all Americans, how much would each citizen pay? Express the answer in decimal notation, rounded to the nearest dollar. 117. Use scientific notation to compute the amount of money that the motion picture industry made from box-office receipts in 2006. Express the answer in scientific notation. 118. Use scientific notation to compute the amount of money that the motion picture industry made from box office receipts in 2005. Express the answer in scientific notation. 119. The mass of one oxygen molecule is 5.3 * 10-23 gram. Find the mass of 20,000 molecules of oxygen. Express the answer in scientific notation. 120. The mass of one hydrogen atom is 1.67 * 10-24 gram. Find the mass of 80,000 hydrogen atoms. Express the answer in scientific notation. 121. In this exercise, use the fact that there are approximately 3.2 * 107 seconds in a year. According to the United States Department of Agriculture, Americans consume 127 chickens per second. How many chickens are eaten per year in the United States? Express the answer in scientific notation. 122. Convert 365 days (one year) to hours, to minutes, and, finally, to seconds, to determine how many seconds there are in a year. Express the answer in scientific notation. Writing in Mathematics 123. Describe what it means to raise a number to a power. In your description, include a discussion of the difference between -52 and 1- 522. 124. Explain the product rule for exponents. Use 2 3 # 2 5 in your explanation. 4 125. Explain the power rule for exponents. Use 1322 in your explanation. 58 126. Explain the quotient rule for exponents. Use 2 in your 5 explanation. 127. Why is 1-3x2212x -52 not simplified? What must be done to simplify the expression? 128. How do you know if a number is written in scientific notation? 30 Chapter P Prerequisites: Fundamental Concepts of Algebra 129. Explain how to convert from scientific to decimal notation and give an example. 130. Explain how to convert from decimal to scientific notation and give an example. Group Exercise Critical Thinking Exercises Make Sense? In Exercises 131–134, determine whether each statement makes sense or does not make sense, and explain your reasoning. 131. There are many exponential expressions that are equal to 2 9 6 36x12, such as 16x62 , 16x3216x92, 361x32 , and 621x22 . -2 132. If 5 is raised to the third power, the result is a number between 0 and 1. 133. The population of Colorado is approximately 4.6 * 1012. 134. I just finished reading a book that contained approximately 1.04 * 105 words. In Exercises 135–142, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 135. 4 -2 6 4 -3 137. 1-22 = 2 4 -4 139. 534.7 = 5.347 * 103 136. 5 -2 138. 5 2 140. 145. Our hearts beat approximately 70 times per minute. Express in scientific notation how many times the heart beats over a lifetime of 80 years. Round the decimal factor in your scientific notation answer to two decimal places. 7 2 #5 -2 8 * 10 -5 7 2 30 4 * 10-5 5 #2 -5 = 2 * 1025 141. 17 * 1052 + 12 * 10-32 = 9 * 102 146. Putting Numbers into Perspective. A large number can be put into perspective by comparing it with another number. For example, we put the $9.2 trillion national debt (Example 6) and the $2.52 trillion the government collected in taxes (Exercise 115) by comparing these numbers to the number of U.S. citizens. For this project, each group member should consult an almanac, a newspaper, or the World Wide Web to find a number greater than one million. Explain to other members of the group the context in which the large number is used. Express the number in scientific notation. Then put the number into perspective by comparing it with another number. Preview Exercises Exercises 147–149 will help you prepare for the material covered in the next section. 147. a. Find 216 # 24. b. Find 216 # 4 . c. Based on your answers to parts (a) and (b), what can you conclude? 143. The mad Dr. Frankenstein has gathered enough bits and pieces (so to speak) for 2 -1 + 2 -2 of his creature-to-be. Write a fraction that represents the amount of his creature that must still be obtained. 148. a. Use a calculator to approximate 2300 to two decimal places. b. Use a calculator to approximate 10 23 to two decimal places. c. Based on your answers to parts (a) and (b), what can you conclude? 144. If bA = MN, bC = M, and bD = N, what is the relationship among A, C, and D? 149. a. Simplify: b. Simplify: 142. 14 * 1032 + 13 * 1022 = 4.3 * 103 Section P.3 Objectives � Evaluate square roots. � Simplify expressions of the form 2a 2 . � Use the product rule to simplify square roots. � Use the quotient rule to simplify square roots. � Add and subtract square roots. � Rationalize denominators. � Evaluate and perform operations with higher roots. � Understand and use rational exponents. 21x + 10x. 2122 + 1022. Radicals and Rational Exponents T his photograph shows mathematical models used by Albert Einstein at a lecture on relativity. Notice the radicals that appear in many of the formulas. Among these models, there is one describing how an astronaut in a moving spaceship ages more slowly than friends who remain on Earth. No description of your world can be complete without roots and radicals. In this section, in addition to reviewing the basics of radical expressions and the use of rational exponents to indicate radicals, you will see how radicals model time dilation for a futuristic highspeed trip to a nearby star. 46 Chapter P Prerequisites: Fundamental Concepts of Algebra Section P.4 Objectives � Understand the vocabulary of polynomials. � Add and subtract polynomials. � Multiply polynomials. � Use FOIL in polynomial multiplication. � Use special products in polynomial multiplication. � Perform operations with polynomials in several variables. Polynomials C an that be Axl, your author’s yellow lab, sharing a special moment with a baby chick? And if it is (it is), what possible relevance can this have to polynomials? An answer is promised before you reach the exercise set. For now, we open the section by defining and describing polynomials. How We Define Polynomials More education results in a higher income. The mathematical models Old Dog Á New Chicks M = - 18x3 + 923x2 - 9603x + 48,446 and W = 17x3 - 450x2 + 6392x - 14,764 describe the median, or middlemost, annual income for men, M, and women, W, who have completed x years of education. We’ll be working with these models and the data upon which they are based in the exercise set. The algebraic expressions that appear on the right sides of the models are examples of polynomials. A polynomial is a single term or the sum of two or more terms containing variables with whole-number exponents. The polynomials above each contain four terms. Equations containing polynomials are used in such diverse areas as science, business, medicine, psychology, and sociology. In this section, we review basic ideas about polynomials and their operations. � Understand the vocabulary of polynomials. How We Describe Polynomials Consider the polynomial 7x3 - 9x2 + 13x - 6. We can express this polynomial as 7x3 + 1- 9x22 + 13x + 1 - 62. The polynomial contains four terms. It is customary to write the terms in the order of descending powers of the variable. This is the standard form of a polynomial. Some polynomials contain only one variable. Each term of such a polynomial in x is of the form axn. If a Z 0, the degree of ax n is n. For example, the degree of the term 7x3 is 3. Study Tip We can express 0 in many ways, including 0x, 0x2, and 0x3. It is impossible to assign a unique exponent to the variable. This is why 0 has no defined degree. The Degree of axn If a Z 0, the degree of axn is n. The degree of a nonzero constant is 0. The constant 0 has no defined degree. Section P.4 Polynomials 47 Here is an example of a polynomial and the degree of each of its four terms: 6x4-3x3+2x-5. degree 4 degree 3 degree 1 degree of nonzero constant: 0 Notice that the exponent on x for the term 2x, meaning 2x1, is understood to be 1. For this reason, the degree of 2x is 1. You can think of - 5 as - 5x0; thus, its degree is 0. A polynomial is simplified when it contains no grouping symbols and no like terms. A simplified polynomial that has exactly one term is called a monomial. A binomial is a simplified polynomial that has two terms. A trinomial is a simplified polynomial with three terms. Simplified polynomials with four or more terms have no special names. The degree of a polynomial is the greatest degree of all the terms of the polynomial. For example, 4x2 + 3x is a binomial of degree 2 because the degree of the first term is 2, and the degree of the other term is less than 2. Also, 7x5 - 2x2 + 4 is a trinomial of degree 5 because the degree of the first term is 5, and the degrees of the other terms are less than 5. Up to now, we have used x to represent the variable in a polynomial. However, any letter can be used. For example, • 7x5 - 3x3 + 8 is a polynomial (in x) of degree 5. Because there are three terms, the polynomial is a trinomial. • 6y3 + 4y2 - y + 3 is a polynomial (in y) of degree 3. Because there are four terms, the polynomial has no special name. • z7 + 22 is a polynomial (in z) of degree 7. Because there are two terms, the polynomial is a binomial. We can tie together the threads of our discussion with the formal definition of a polynomial in one variable. In this definition, the coefficients of the terms are represented by a n (read “a sub n”), an - 1 (read “a sub n minus 1”), an - 2 , and so on. The small letters to the lower right of each a are called subscripts and are not exponents. Subscripts are used to distinguish one constant from another when a large and undetermined number of such constants are needed. Definition of a Polynomial in x A polynomial in x is an algebraic expression of the form anxn + an - 1xn - 1 + an - 2xn - 2 + Á + a1x + a0 , where an , an - 1 , an - 2 , Á , a1 , and a0 are real numbers, an Z 0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and a 0 is the constant term. � Add and subtract polynomials. Adding and Subtracting Polynomials Polynomials are added and subtracted by combining like terms. For example, we can combine the monomials -9x3 and 13x3 using addition as follows: –9x3+13x3=(–9+13)x3=4x3. These like terms both contain x to the third power. Add coefficients and keep the same variable factor, x3. 48 Chapter P Prerequisites: Fundamental Concepts of Algebra Adding and Subtracting Polynomials EXAMPLE 1 Perform the indicated operations and simplify: a. 1 -9x3 + 7x2 - 5x + 32 + 113x3 + 2x2 - 8x - 62 b. 17x3 - 8x2 + 9x - 62 - 12x3 - 6x2 - 3x + 92. Solution a. 1-9x3 + 7x2 - 5x + 32 + 113x3 + 2x2 - 8x - 62 = 1 -9x3 + 13x32 + 17x2 + 2x22 + 1-5x - 8x2 + 13 - 62 Group like terms. = 4x3 + 9x2 + 1-13x2 + 1 - 32 3 2 3 2 Combine like terms. = 4x + 9x - 13x - 3 Study Tip b. 2 7x - 8x + 9x - 6 -2x3 + 6x2 + 3x - 9 5x3 - 2x2 + 12x - 15 The like terms can be combined by adding their coefficients and keeping the same variable factor. 2 (7x -8x +9x-6)-(2x -6x -3x+9) You can also arrange like terms in columns and combine vertically: 3 Simplify. 3 Change the sign of each coefficient. =(7x3-8x2+9x-6)+(–2x3+6x2+3x-9) Rewrite subtraction as addition of the additive inverse. = 17x3 - 2x32 + 1-8x2 + 6x22 + 19x + 3x2 + 1- 6 - 92 Group like terms. = 5x3 + 1-2x22 + 12x + 1-152 3 Combine like terms. 2 = 5x - 2x + 12x - 15 Check Point 1 Simplify. Perform the indicated operations and simplify: a. 1 -17x3 + 4x2 - 11x - 52 + 116x3 - 3x2 + 3x - 152 b. 113x3 - 9x2 - 7x + 12 - 1 -7x3 + 2x2 - 5x + 92. � Multiply polynomials. Multiplying Polynomials The product of two monomials is obtained by using properties of exponents. For example, Study Tip Don’t confuse adding and multiplying monomials. (–8x6)(5x3)=–8 ⴢ 5x6±3=–40x9. Addition: Multiply coefficients and add exponents. 5x4 + 6x4 = 11x4 Multiplication: 15x4216x42 = 15 # 621x4 # x42 Furthermore, we can use the distributive property to multiply a monomial and a polynomial that is not a monomial. For example, = 30x4 + 4 = 30x8 Only like terms can be added or subtracted, but unlike terms may be multiplied. 3x4(2x3-7x+3)=3x4 ⴢ 2x3-3x4 ⴢ 7x+3x4 ⴢ 3=6x7-21x5+9x4. Monomial Trinomial Addition: 5x4 + 3x2 cannot be simplified. Multiplication: 15x4213x22 = 15 # 321x4 # x22 How do we multiply two polynomials if neither is a monomial? For example, consider (2x+3)(x2+4x +5). = 15x4 + 2 = 15x6 Binomial Trinomial Section P.4 Polynomials 49 One way to perform (2x + 3)(x2 + 4x + 5) is to distribute 2x throughout the trinomial 2x1x2 + 4x + 52 and 3 throughout the trinomial 31x2 + 4x + 52. Then combine the like terms that result. Multiplying Polynomials When Neither Is a Monomial Multiply each term of one polynomial by each term of the other polynomial. Then combine like terms. Multiplying a Binomial and a Trinomial EXAMPLE 2 Multiply: Solution 12x + 321x2 + 4x + 52. 12x + 321x2 + 4x + 52 = 2x1x2 + 4x + 52 + 31x2 + 4x + 52 = 2x # x2 + 2x # 4x + 2x # 5 + 3x2 + 3 # 4x + 3 # 5 3 2 2 Multiply the trinomial by each term of the binomial. Use the distributive property. = 2x + 8x + 10x + 3x + 12x + 15 Multiply monomials: Multiply coefficients and add exponents. = 2x3 + 11x2 + 22x + 15 Combine like terms: 8x2 + 3x2 = 11x2 and 10x + 12x = 22x. Another method for performing the multiplication is to use a vertical format similar to that used for multiplying whole numbers. x2+4x+ 5 2x+ 3 3x2+12x+15 Write like terms in the same column. 2x3+ 8x2+10x Combine like terms. Check Point � Use FOIL in polynomial multiplication. 2 3 3(x2 + 4x + 5) 2x(x2 + 4x + 5) 2 2x +11x +22x+15 Multiply: 15x - 2213x2 - 5x + 42. The Product of Two Binomials: FOIL Frequently, we need to find the product of two binomials. One way to perform this multiplication is to distribute each term in the first binomial through the second binomial. For example, we can find the product of the binomials 3x + 2 and 4x + 5 as follows: (3x+2)(4x+5)=3x(4x+5)+2(4x+5) =3x(4x)+3x(5)+2(4x)+2(5) Distribute 3x Distribute 2 =12x2+15x+8x+10. over 4x + 5. over 4x + 5. We'll combine these like terms later. For now, our interest is in how to obtain each of these four terms. 50 Chapter P Prerequisites: Fundamental Concepts of Algebra We can also find the product of 3x + 2 and 4x + 5 using a method called FOIL, which is based on our work shown at the bottom of the previous page. Any two binomials can be quickly multiplied by using the FOIL method, in which F represents the product of the first terms in each binomial, O represents the product of the outside terms, I represents the product of the inside terms, and L represents the product of the last, or second, terms in each binomial. For example, we can use the FOIL method to find the product of the binomials 3x + 2 and 4x + 5 as follows: last first F O I L (3x+2)(4x+5)=12x2+15x+8x+10 inside outside Product of First terms Product of Outside terms Product of Inside terms Product of Last terms =12x2+23x+10 Combine like terms. In general, here’s how to use the FOIL method to find the product of ax + b and cx + d: Using the FOIL Method to Multiply Binomials last first F O I L (ax+b)(cx+d)=ax ⴢ cx+ax ⴢ d+b ⴢ cx+b ⴢ d inside outside Solution Product of Outside terms Product of Inside terms Product of Last terms Using the FOIL Method EXAMPLE 3 Multiply: Product of First terms 13x + 4215x - 32. last first F O I L (3x+4)(5x-3)=3x ⴢ 5x+3x(–3)+4 ⴢ 5x+4(–3) inside outside Check Point � Use special products in polynomial multiplication. 3 Multiply: =15x2-9x+20x-12 =15x2+11x-12 Combine like terms. 17x - 5214x - 32. Special Products There are several products that occur so frequently that it’s convenient to memorize the form, or pattern, of these formulas. Section P.4 Polynomials 51 Special Products Let A and B represent real numbers, variables, or algebraic expressions. Study Tip Although it’s convenient to memorize these forms, the FOIL method can be used on all five examples in the box. To cube x + 4, you can first square x + 4 using FOIL and then multiply this result by x + 4. In short, you do not necessarily have to utilize these special formulas. What is the advantage of knowing and using these forms? Special Product Example Sum and Difference of Two Terms 1A + B21A - B2 = A2 - B2 12x + 3212x - 32 = 12x22 - 32 = 4x2 - 9 Squaring a Binomial 1A + B22 = A2 + 2AB + B2 1y + 522 = y2 + 2 # y # 5 + 52 1A - B22 = A2 - 2AB + B2 13x - 422 = y2 + 10y + 25 = 13x22 - 2 # 3x # 4 + 4 2 = 9x2 - 24x + 16 Cubing a Binomial 1A + B23 = A3 + 3A2B + 3AB2 + B3 1x + 423 = x3 + 3x2142 + 3x1422 + 4 3 = x3 + 12x2 + 48x + 64 1A - B23 = A3 - 3A2B + 3AB2 - B3 1x - 223 = x3 - 3x2122 + 3x1222 - 2 3 = x3 - 6x2 + 12x - 8 � Perform operations with polynomials in several variables. Polynomials in Several Variables A polynomial in two variables, x and y, contains the sum of one or more monomials in the form axnym. The constant, a, is the coefficient. The exponents, n and m, represent whole numbers. The degree of the monomial axnym is n + m. Here is an example of a polynomial in two variables: The coefficients are 7, −17, 1, −6, and 9. 7x2y3 Degree of monomial: 2+3=5 - 17x4y2 Degree of monomial: 4+2=6 + xy Degree of monomial A1x1y1B: 1+1=2 - 6y2 + Degree of monomial A−6x0y2B: 0+2=2 9. Degree of monomial A9x0y0B: 0+0=0 The degree of a polynomial in two variables is the highest degree of all its terms. For the preceding polynomial, the degree is 6. Polynomials containing two or more variables can be added, subtracted, and multiplied just like polynomials that contain only one variable. For example, we can add the monomials -7xy2 and 13xy2 as follows: –7xy2+13xy2=(–7+13)xy2=6xy2. These like terms both contain the variable factors x and y2. Add coefficients and keep the same variable factors, xy2. 52 Chapter P Prerequisites: Fundamental Concepts of Algebra Multiplying Polynomials in Two Variables EXAMPLE 4 b. 15x + 3y22. Multiply: a. 1x + 4y213x - 5y2 Solution We will perform the multiplication in part (a) using the FOIL method. We will multiply in part (b) using the formula for the square of a binomial sum, 1A + B22. a. (x+4y)(3x-5y) F Multiply these binomials using the FOIL method. O I L =(x)(3x)+(x)(–5y)+(4y)(3x)+(4y)(–5y) =3x2-5xy+12xy-20y2 =3x2+7xy-20y2 Combine like terms. = (A + B)2 A2 + 2 ⴢ A ⴢ B + B2 b. (5x+3y)2=(5x)2+2(5x)(3y)+(3y)2 =25x2+30xy+9y2 Check Point 4 Multiply: a. 17x - 6y213x - y2 b. 12x + 4y22. Special products can sometimes be used to find the products of certain trinomials, as illustrated in Example 5. Using the Special Products EXAMPLE 5 Multiply: a. 17x + 5 + 4y217x + 5 - 4y2 b. 13x + y + 122. Solution a. By grouping the first two terms within each of the parentheses, we can find the product using the form for the sum and difference of two terms. + (A B) (A ⴢ − = B) A2 − B2 [(7x+5)+4y] ⴢ [(7x+5)-4y]=(7x+5)2-(4y)2 = 17x22 + 2 # 7x # 5 + 52 - 14y22 = 49x2 + 70x + 25 - 16y2 b. We can group the terms of 13x + y + 12 so that the formula for the square of a binomial can be applied. 2 (A + B)2 = A2 + 2 ⴢ A ⴢ B + B2 [(3x+y)+1]2 =(3x+y)2+2 ⴢ (3x+y) ⴢ 1+12 = 9x2 + 6xy + y2 + 6x + 2y + 1 Check Point 5 Multiply: a. 13x + 2 + 5y213x + 2 - 5y2 b. 12x + y + 322. Section P.4 Polynomials 53 Labrador Retrievers and Polynomial Multiplication The color of a Labrador retriever is determined by its pair of genes. A single gene is inherited at random from each parent. The black-fur gene, B, is dominant. The yellow-fur gene, Y, is recessive. This means that labs with at least one black-fur gene (BB or BY) have black coats. Only labs with two yellow-fur genes (YY) have yellow coats. Axl, your author’s yellow lab, inherited his genetic makeup from two black BY parents. Second BY parent, a black lab with a recessive yellow-fur gene First BY parent, a black lab with a recessive yellow-fur gene B Y B BB BY Y BY YY The table shows the four possible combinations of color genes that BY parents can pass to their offspring. Because YY is one of four possible outcomes, the probability that a yellow lab like Axl will be the offspring of these black parents is 14 . The probabilities suggested by the table can be modeled by the expression A 12 B + 12 Y B 2. 2 2 1 1 1 1 1 1 a B+ Yb = a Bb +2 a Bb a Yb + a Yb 2 2 2 2 2 2 1 BB + 4 = The probability of a black lab with two dominant black genes is 1 . 4 1 BY 2 2 1 + YY 4 The probability of a black lab with a recessive yellow gene is 1 . 2 The probability of a yellow lab with two recessive yellow genes is 1 . 4 Exercise Set P.4 Practice Exercises In Exercises 15–82, find each product. In Exercises 1–4, is the algebraic expression a polynomial? If it is, write the polynomial in standard form. 2 2. 2x + 3x 1. 2x + 3x - 5 2x + 3 3. x 2 -1 - 5 3 4 4. x - x + x - 5 In Exercises 5–8, find the degree of the polynomial. 2 3 6. -4x + 7x - 11 5. 3x - 5x + 4 2 3 2 4 7. x - 4x + 9x - 12x + 63 8. x2 - 8x3 + 15x4 + 91 In Exercises 9–14, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree. 9. 1-6x3 + 5x2 - 8x + 92 + 117x3 + 2x2 - 4x - 132 10. 1-7x3 + 6x2 - 11x + 132 + 119x3 - 11x2 + 7x - 172 11. 117x3 - 5x2 + 4x - 32 - 15x3 - 9x2 - 8x + 112 12. 118x4 - 2x3 - 7x + 82 - 19x4 - 6x3 - 5x + 72 13. 15x2 - 7x - 82 + 12x2 - 3x + 72 - 1x2 - 4x - 32 14. 18x2 + 7x - 52 - 13x2 - 4x2 - 1- 6x3 - 5x2 + 32 15. 1x + 121x2 - x + 12 16. 1x + 521x2 - 5x + 252 19. 1x + 721x + 32 20. 1x + 821x + 52 17. 12x - 321x2 - 3x + 52 21. 1x - 521x + 32 23. 13x + 5212x + 12 25. 12x - 3215x + 32 27. 15x - 4213x - 72 2 2 29. 18x3 + 321x2 - 52 31. 1x + 321x - 32 33. 13x + 2213x - 22 35. 15 - 7x215 + 7x2 37. 14x2 + 5x214x2 - 5x2 39. 11 - y5211 + y52 41. 1x + 222 18. 12x - 121x2 - 4x + 32 22. 1x - 121x + 22 24. 17x + 4213x + 12 26. 12x - 5217x + 22 28. 17x2 - 2213x2 - 52 30. 17x3 + 521x2 - 22 32. 1x + 521x - 52 34. 12x + 5212x - 52 36. 14 - 3x214 + 3x2 38. 13x2 + 4x213x2 - 4x2 40. 12 - y5212 + y52 42. 1x + 522 54 Chapter P Prerequisites: Fundamental Concepts of Algebra 44. 13x + 222 46. 1x - 422 48. 15x2 - 32 2 50. 19 - 5x22 Men 52. 1x + 22 56. 1x - 123 57. 13x - 42 3 58. 12x - 32 3 59. 1x + 5y217x + 3y2 61. 1x - 3y212x + 7y2 63. 13xy - 1215xy + 22 65. 17x + 5y22 67. 1x2y2 - 32 2 69. 1x - y21x2 + xy + y22 71. 13x + 5y213x - 5y2 73. 1x + y + 321x + y - 32 60. 1x + 9y216x + 7y2 62. 13x - y212x + 5y2 64. 17x2 y + 1212x2 y - 32 66. 19x + 7y22 68. 1x2 y 2 - 52 2 70. 1x + y21x2 - xy + y22 72. 17x + 3y217x - 3y2 75. 13x + 7 - 5y213x + 7 + 5y2 78. 38y + 17 - 3x2438y - 17 - 3x24 79. 1x + y + 122 80. 1x + y + 222 81. 12x + y + 122 82. 15x + 1 + 6y22 Practice Plus In Exercises 83–90, perform the indicated operation or operations. 83. 13x + 4y22 - 13x - 4y22 84. 15x + 2y22 - 15x - 2y22 85. 15x - 7213x - 22 - 14x - 5216x - 12 86. 13x + 5212x - 92 - 17x - 221x - 12 87. 12x + 5212x - 5214x2 + 252 88. 13x + 4213x - 4219x2 + 162 12x - 72 $40 $30 $20 $10 8 10 12 13 14 16 Years of School Completed 18 20 Source: Bureau of the Census Here are polynomial models that describe the median annual income for men, M, and for women, W, who have completed x years of education: M = 177x2 + 288x + 7075 W = 255x2 - 2956x + 24,336 91. a. Use the equation defined by a polynomial of degree 2 to find the median annual income for a man with 16 years of education. Does this underestimate or overestimate the median income shown by the bar graph? By how much? b. Use the equations defined by polynomials of degree 3 to find a mathematical model for M - W. c. According to the model in part (b), what is the difference in the median annual income between men and women with 14 years of education? d. According to the data displayed by the graph, what is the actual difference in the median annual income between men and women with 14 years of education? Did the result of part (c) underestimate or overestimate this difference? By how much? 77. 35y - 12x + 32435y + 12x + 324 3 $50 Exercises 91–92 are based on these models and the data displayed by the graph. 76. 15x + 7y - 2215x + 7y + 22 12x - 725 $60 M = - 18x3 + 923x2 - 9603x + 48,446 W = 17x3 - 450x2 + 6392x - 14,764. 74. 1x + y + 521x + y - 52 89. $70 82,401 68,875 55. 1x - 323 $80 71,530 54. 13x + 423 51,316 53. 12x + 323 Women $90 3 Median Annual Income (thousands of dollars) 3 57,220 41,681 51. 1x + 12 Median Annual Income, by Level of Education, 2004 44,404 33,481 49. 17 - 2x22 41,895 30,816 2 35,725 26,029 47. 14x2 - 12 As you complete more years of education, you can count on a greater income. The bar graph shows the median, or middlemost, annual income for Americans, by level of education, in 2004. 26,277 19,169 45. 1x - 322 Application Exercises 21,659 17,023 43. 12x + 322 90. 15x - 326 15x - 324 92. a. Use the equation defined by a polynomial of degree 2 to find the median annual income for a woman with 18 years of education. Does this underestimate or overestimate the median income shown by the bar graph? By how much? b. Use the equations defined by polynomials of degree 3 to find a mathematical model for M - W. c. According to the model in part (b), what is the difference in the median annual income between men and women with 16 years of education? d. According to the data displayed by the graph, what is the actual difference in the median annual income between men and women with 16 years of education? Did the result of part (c) underestimate or overestimate this difference? By how much? Section P.4 Polynomials The volume, V, of a rectangular solid with length l, width w, and height h is given by the formula V = lwh. In Exercises 93–94, use this formula to write a polynomial in standard form that models, or represents, the volume of the open box. 93. x 8 − 2x 10 − 2x 55 104. I used the FOIL method to find the product of x + 5 and x2 + 2x + 1. 105. Many English words have prefixes with meanings similar to those used to describe polynomials, such as monologue, binocular, and tricuspid. 106. Special-product formulas have patterns that make their multiplications quicker than using the FOIL method. 107. Express the area of the plane figure shown as a polynomial in standard form. 94. x x 8 − 2x x x−1 5 − 2x x+3 In Exercises 95–96, write a polynomial in standard form that models, or represents, the area of the shaded region. 95. x+9 x+3 x+5 x+1 In Exercises 108–109, represent the volume of each figure as a polynomial in standard form. 108. x 96. x+4 x+3 x+1 2x − 1 x+2 x+1 x x+3 109. Writing in Mathematics 97. What is a polynomial in x? 98. Explain how to subtract polynomials. 99. Explain how to multiply two binomials using the FOIL method. Give an example with your explanation. 100. Explain how to find the product of the sum and difference of two terms. Give an example with your explanation. 101. Explain how to square a binomial difference. Give an example with your explanation. x+2 3 x x+5 2x + 1 110. Simplify: 1yn + 221yn - 22 - 1yn - 32 . 2 102. Explain how to find the degree of a polynomial in two variables. Preview Exercises Critical Thinking Exercises Exercises 111–113 will help you prepare for the material covered in the next section. In each exercise, replace the boxed question mark with an integer that results in the given product. Some trial and error may be necessary. Make Sense? In Exercises 103–106, determine whether each statement makes sense or does not make sense, and explain your reasoning. 103. Knowing the difference between factors and terms is 2 important: In 13x2 y2 , I can distribute the exponent 2 on 2 each factor, but in 13x2 + y2 , I cannot do the same thing on each term. 111. 1x + 321x + 冷 ? 冷2 = x2 + 7x + 12 112. 1x - 冷 ? 冷21x - 122 = x2 - 14x + 24 113. 14x + 1212x - 冷 ? 冷2 = 8x2 - 10x - 3 82 Chapter P Prerequisites: Fundamental Concepts of Algebra 119. 1x - y2-1 + 1x - y2-2 111. When performing the division 1x + 32 7x , , x + 3 x - 5 2 I began by dividing the numerator and the denominator by the common factor, x + 3. 112. I subtracted 3x - 5 x - 3 from and obtained a constant. x - 1 x - 1 In Exercises 113–116, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. x2 - 25 = x - 5 113. x - 5 - 3y - 6 114. The expression simplifies to the consecutive y + 2 integer that follows - 4. 2x - 1 3x - 1 5x - 2 + = 0 x - 7 x - 7 x - 7 1 7 = 116. 6 + x x 115. 1 1 1 - n - 2n xn - 1 x + 1 x - 1 1 1 1 1 b a1 b a1 b 118. a 1 - b a1 x x + 1 x + 2 x + 3 Objectives � Solve linear equations in � � � � � � � � 쐅 쐈 쐉 one variable. Solve linear equations containing fractions. Solve rational equations with variables in the denominators. Solve a formula for a variable. Solve equations involving absolute value. Solve quadratic equations by factoring. Solve quadratic equations by the square root property. Solve quadratic equations by completing the square. Solve quadratic equations using the quadratic formula. Use the discriminant to determine the number and type of solutions of quadratic equations. Determine the most efficient method to use when solving a quadratic equation. Solve radical equations. Preview Exercises Exercises 121–123 will help you prepare for the material covered in the next section. 121. If 6 is substituted for x in the equation 2(x - 3) - 17 = 13 - 3 (x + 2), 122. Multiply and simplify: 117. P.7 does to each number x. is the resulting statement true or false? In Exercises 117–119, perform the indicated operations. Section 120. In one short sentence, five words or less, explain what 1 1 1 + 2 + 3 x x x 1 1 1 + 5 + 6 x4 x x 12a x + 2 x - 1 b. 4 3 - b - 2b2 - 4ac 2a for a = 2, b = 9, and c = - 5. 123. Evaluate Equations M ath tattoos. Who knew? Do you recognize the significance of this tattoo? The algebraic expression gives the solutions of a quadratic equation. In this section, we will review how to solve a variety of equations, including linear equations, quadratic equations, and radical equations. Solving Linear Equations in One Variable We begin with a general definition of a linear equation in one variable. Definition of a Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0, where a and b are real numbers, and a Z 0. An example of a linear equation in one variable is 4x + 12 = 0. Section P.7 Equations 83 Solving an equation in x involves determining all values of x that result in a true statement when substituted into the equation. Such values are solutions, or roots, of the equation. For example, substitute - 3 for x in 4x + 12 = 0. We obtain 41 - 32 + 12 = 0, or -12 + 12 = 0. This simplifies to the true statement 0 = 0. Thus, -3 is a solution of the equation 4x + 12 = 0. We also say that -3 satisfies the equation 4x + 12 = 0, because when we substitute - 3 for x, a true statement results. The set of all such solutions is called the equation’s solution set. For example, the solution set of the equation 4x + 12 = 0 is 5 -36 because -3 is the equation’s only solution. Two or more equations that have the same solution set are called equivalent equations. For example, the equations 4x + 12 = 0 and 4x = - 12 and x = - 3 are equivalent equations because the solution set for each is 5 -36. To solve a linear equation in x, we transform the equation into an equivalent equation one or more times. Our final equivalent equation should be of the form x = a number. The solution set of this equation is the set consisting of the number. To generate equivalent equations, we will use the following principles: Generating Equivalent Equations An equation can be transformed into an equivalent equation by one or more of the following operations: Example 1. Simplify an expression by removing grouping symbols and combining like terms. • 3(x-6)=6x-x 3x - 18 = 5x 3x - 18 = 5x 2. Add (or subtract) the same real number or variable expression on both sides of the equation. • 3. Multiply (or divide) by the same nonzero quantity on both sides of the equation. • –18=2x 2x –18 = 2 2 -9 = x 4. Interchange the two sides of the equation. • -9 = x x = -9 3x-18-3x=5x-3x Subtract 3x from both sides of the equation. -18 = 2x Divide both sides of the equation by 2. If you look closely at the equations in the box, you will notice that we have solved the equation 31x - 62 = 6x - x. The final equation, x = - 9, with x isolated on the left side, shows that 5- 96 is the solution set. The idea in solving a linear equation is to get the variable by itself on one side of the equal sign and a number by itself on the other side. 84 Chapter P Prerequisites: Fundamental Concepts of Algebra � Solve linear equations in one variable. Here is a step-by-step procedure for solving a linear equation in one variable. Not all of these steps are necessary to solve every equation. Solving a Linear Equation 1. Simplify the algebraic expression on each side by removing grouping symbols and combining like terms. 2. Collect all the variable terms on one side and all the numbers, or constant terms, on the other side. 3. Isolate the variable and solve. 4. Check the proposed solution in the original equation. Solving a Linear Equation EXAMPLE 1 21x - 32 - 17 = 13 - 31x + 22. Solve and check: Solution Step 1 Simplify the algebraic expression on each side. Do not begin with 13 – 3. Multiplication (the distributive property) is applied before subtraction. 2(x-3)-17=13-3(x+2) 2x - 6 - 17 = 13 - 3x - 6 2x - 23 = - 3x + 7 This is the given equation. Use the distributive property. Combine like terms. Step 2 Collect variable terms on one side and constant terms on the other side. We will collect variable terms on the left by adding 3x to both sides. We will collect the numbers on the right by adding 23 to both sides. 2x - 23 + 3x = - 3x + 7 + 3x 5x - 23 = 7 Add 3x to both sides. Simplify: 2x + 3x = 5x. 5x - 23 + 23 = 7 + 23 5x = 30 Add 23 to both sides. Simplify. Step 3 Isolate the variable and solve. We isolate the variable, x, by dividing both sides of 5x = 30 by 5. 5x 30 = 5 5 x = 6 Divide both sides by 5. Simplify. Step 4 Check the proposed solution in the original equation. Substitute 6 for x in the original equation. Discovery Solve the equation in Example 1 by collecting terms with the variable on the right and numerical terms on the left. What do you observe? 21x - 32 - 17 = 13 - 31x + 22 This is the original equation. 216 - 32 - 17 ⱨ 13 - 316 + 22 Substitute 6 for x. 2132 - 17 ⱨ 13 - 3182 Simplify inside parentheses. 6 - 17 ⱨ 13 - 24 Multiply. -11 = - 11 Subtract. The true statement - 11 = - 11 verifies that the solution set is 566. Check Point 1 Solve and check: 412x + 12 = 29 + 312x - 52. Section P.7 Equations 85 � Solve linear equations containing fractions. Linear Equations with Fractions Equations are easier to solve when they do not contain fractions. How do we remove fractions from an equation? We begin by multiplying both sides of the equation by the least common denominator of any fractions in the equation. The least common denominator is the smallest number that all denominators will divide into. Multiplying every term on both sides of the equation by the least common denominator will eliminate the fractions in the equation. Example 2 shows how we “clear an equation of fractions.” Solving a Linear Equation Involving Fractions EXAMPLE 2 x + 2 x - 1 = 2. 4 3 Solve and check: Solution The fractional terms have denominators of 4 and 3. The smallest number that is divisible by 4 and 3 is 12. We begin by multiplying both sides of the equation by 12, the least common denominator. x + 2 x - 1 = 2 4 3 x+2 x-1 12 a b =12 ⴢ 2 4 3 This is the given equation. Multiply both sides by 12. 12 a x + 2 x - 1 b - 12 a b = 24 4 3 Use the distributive property and multiply each term on the left by 12. 12 ¢ 4 x + 2 x - 1 ≤ - 12 ¢ ≤ = 24 4 3 Divide out common factors in each multiplication on the left. 3 1 1 31x + 22 - 41x - 12 = 24 The fractions are now cleared. 3x + 6 - 4x + 4 = 24 Use the distributive property. -x + 10 = 24 -x + 10 - 10 = 24 - 10 –x=14 Combine like terms: 3x - 4x = - x and 6 + 4 = 10. Subtract 10 from both sides. Simplify. We’re not finished. A negative sign should not precede the variable. Isolate x by multiplying or dividing both sides of this equation by -1. -x 14 = -1 -1 x = - 14 Divide both sides by ⴚ 1. Simplify. Check the proposed solution. Substitute - 14 for x in the original equation. You should obtain 2 = 2. This true statement verifies that the solution set is 5-146. Check Point 2 Solve and check: x - 3 5 x + 5 = . 4 14 7 86 Chapter P Prerequisites: Fundamental Concepts of Algebra � Rational Equations Solve rational equations with variables in the denominators. A rational equation is an equation containing one or more rational expressions. In Example 2, we solved a rational equation with constants in the denominators. This rational equation was a linear equation. Now, let’s consider a rational equation such as 3 1 4 + = 2 . x + 6 x - 2 x + 4x - 12 Can you see how this rational equation differs from the rational equation that we solved earlier? The variable appears in the denominators. Although this rational equation is not a linear equation, the solution procedure still involves multiplying each side by the least common denominator. However, we must avoid any values of the variable that make a denominator zero. Solving a Rational Equation EXAMPLE 3 Solve: 3 1 4 . + = 2 x + 6 x - 2 x + 4x - 12 Solution To identify values of x that make denominators zero, let’s factor x2 + 4x - 12, the denominator on the right. This factorization is also necessary in identifying the least common denominator. 1 4 3 + = x+6 x-2 (x+6)(x-2) This denominator is zero if x =−6. This denominator is zero if x =2. This denominator is zero if x =−6 or x =2. We see that x cannot equal -6 or 2. The least common denominator is 1x + 621x - 22. 3 1 4 + = , x Z - 6, x Z 2 x + 6 x - 2 1x + 621x - 22 (x+6)(x-2) a 1x + 62 1x - 22 # 3 1 4 + b =(x+6)(x-2) ⴢ x+6 x-2 (x+6)(x-2) This is the given equation with a denominator factored. Multiply both sides by 1x + 621x - 22, the LCD. Use the distributive property 3 1 4 + 1x + 62 1x - 22 # = 1x + 62 1x - 22 # x + 6 x - 2 1x + 62 1x - 22 and divide out common factors. 31x - 22 + 11x + 62 = 4 3x - 6 + x + 6 = 4 Simplify. This equation is cleared of fractions. Use the distributive property. 4x = 4 Combine like terms. 4x 4 = 4 4 Divide both sides by 4. x = 1 Simplify. This is not part of the restriction that x Z - 6 and x Z 2. Check the proposed solution. Substitute 1 for x in the original equation. You should obtain - 47 = - 47 . This true statement verifies that the solution set is 516. Check Point 3 Solve: 6 5 -20 . = 2 x + 3 x - 2 x + x - 6 Section P.7 Equations 87 Solving a Rational Equation EXAMPLE 4 Solve: 1 1 2 = 2 . x + 1 x - 1 x - 1 Solution We begin by factoring x2 - 1. 1 2 1 = x+1 (x+1)(x-1) x-1 This denominator is zero if x = −1. This denominator is zero if x = −1 or x = 1. This denominator is zero if x = 1. We see that x cannot equal - 1 or 1. The least common denominator is 1x + 121x - 12. 1 2 1 = , x Z - 1, x Z 1 x + 1 1x + 121x - 12 x - 1 This is the given equation with a denominator factored. (x+1)(x-1) ⴢ 1 2 1 =(x+1)(x-1) a b x+1 (x+1)(x-1) x-1 1x + 12 1x - 12 # 1 2 1 and divide out common = 1x + 12 1x - 12 # - 1x + 12 1x - 12 # x + 1 1x + 12 1x - 12 1x - 12 factors. Use the distributive property 11x - 12 = 2 - 1x + 12 Simplify. This equation is cleared of fractions. x - 1 = 2 - x - 1 Simplify. x - 1 = -x + 1 Combine numerical terms. x + x - 1 = -x + x + 1 Add x to both sides. 2x - 1 = 1 Simplify. 2x - 1 + 1 = 1 + 1 Add 1 to both sides. 2x = 2 2x 2 = 2 2 x = 1 Study Tip Reject any proposed solution that causes any denominator in an equation to equal 0. Simplify. Divide both sides by 2. Simplify. The proposed solution, 1, is not a solution because of the restriction that x Z 1. There is no solution to this equation. The solution set for this equation contains no elements. The solution set is ⭋, the empty set. Check Point � Multiply both sides by 1x + 121x - 12, the LCD. Solve a formula for a variable. 4 Solve: 1 4 1 = 2 . x + 2 x 2 x - 4 Solving a Formula for One of Its Variables Solving a formula for a variable means rewriting the formula so that the variable is isolated on one side of the equation. It does not mean obtaining a numerical value for that variable. To solve a formula for one of its variables, treat that variable as if it were the only variable in the equation. Think of the other variables as if they were numbers. 88 Chapter P Prerequisites: Fundamental Concepts of Algebra Solving a Formula for a Variable EXAMPLE 5 If you wear glasses, did you know that each lens has a measurement called its focal length, f? When an object is in focus, its distance from the lens, p, and the distance from the lens to your retina, q, satisfy the formula p q 1 1 1 + = . p q f Figure P.12 (See Figure P.12.) Solve this formula for p. Solution Our goal is to isolate the variable p. We begin by multiplying both sides by the least common denominator, pqf, to clear the equation of fractions. 1 1 1 + = q p f We need to isolate p. pqf a p qfa This is the given formula. 1 1 1 + b =pqf a b q p f 1 1 1 b + p q fa b = pq f a b p q f qf+pf=pq Multiply both sides by pqf, the LCD. Use the distributive property on the left side and divide out common factors. Simplify. The formula is cleared of fractions. We need to isolate p. Study Tip You cannot solve qf + pf = pq for p by dividing both sides by q and writing qf + pf = p. q To collect terms with p on one side of the equation, subtract pf from both sides. Then factor p from the two resulting terms on the right to convert two occurrences of p into one. qf + pf = pq qf + pf - pf = pq - pf When a formula is solved for a specified variable, that variable must be isolated on one side. The variable p occurs on both sides of Check Point Factor out p, the specified variable. Solve for q: Divide both sides by q - f and solve for p. Simplify. 1 1 1 + = . p q f ƒ x ƒ = 2. x 0 1 兩2兩 = 2 Figure P.13 qf = p1q - f2 We have seen that the absolute value of x, denoted ƒ x ƒ , describes the distance of x from zero on a number line. Now consider an absolute value equation, such as 兩−2兩 = 2 −1 Simplify. Equations Involving Absolute Value Solve equations involving absolute value. −2 5 Subtract pf from both sides. qf = pq - pf p1q - f2 qf = q - f q - f qf = p q - f qf + pf = p. q � This is the equation cleared of fractions. 2 This means that we must determine real numbers whose distance from the origin on a number line is 2. Figure P.13 shows that there are two numbers such that ƒ x ƒ = 2, namely, 2 and -2. We write x = 2 or x = - 2. This observation can be generalized as follows: Section P.7 Equations 89 Rewriting an Absolute Value Equation without Absolute Value Bars If c is a positive real number and u represents any algebraic expression, then ƒ u ƒ = c is equivalent to u = c or u = - c. Solving an Equation Involving Absolute Value EXAMPLE 6 Solve: 5 ƒ 1 - 4x ƒ - 15 = 0. Solution 5|1-4x|-15=0 This is the given equation. We need to isolate 兩1 − 4x兩, the absolute value expression. 5 ƒ 1 - 4x ƒ = 15 ƒ 1 - 4x ƒ = 3 1 - 4x = 3 or 1 - 4x = - 3 - 4x = 2 - 4x = - 4 1 x = - 2 x = 1 Add 15 to both sides. Divide both sides by 5. Rewrite ƒ u ƒ = c as u = c or u = - c. Subtract 1 from both sides of each equation. Divide both sides of each equation by -4. Take a moment to check - 12 and 1, the proposed solutions, in the original equation, 5 ƒ 1 - 4x ƒ - 15 = 0. In each case, you should obtain the true statement 0 = 0. The solution set is E - 12 , 1 F . Check Point 6 Solve: 4 ƒ 1 - 2x ƒ - 20 = 0. The absolute value of a number is never negative. Thus, if u is an algebraic expression and c is a negative number, then ƒ u ƒ = c has no solution. For example, the equation ƒ 3x - 6 ƒ = - 2 has no solution because ƒ 3x - 6 ƒ cannot be negative. The solution set is ⭋, the empty set. The absolute value of 0 is 0. Thus, if u is an algebraic expression and ƒ u ƒ = 0, the solution is found by solving u = 0. For example, the solution of ƒ x - 2 ƒ = 0 is obtained by solving x - 2 = 0. The solution is 2 and the solution set is 526. � Solve quadratic equations by factoring. Quadratic Equations and Factoring Linear equations are first-degree polynomial equations of the form ax + b = 0. Quadratic equations are second-degree polynomial equations and contain an additional term involving the square of the variable. Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the general form ax2 + bx + c = 0, where a, b, and c are real numbers, with a Z 0. A quadratic equation in x is also called a second-degree polynomial equation in x. Here are examples of quadratic equations in general form: 4x2-2x a=4 b = −2 =0 c=0 2x2+7x-4=0. a=2 b=7 c = −4 Some quadratic equations, including the two shown above, can be solved by factoring and using the zero-product principle. 90 Chapter P Prerequisites: Fundamental Concepts of Algebra The Zero-Product Principle If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB = 0, then A = 0 or B = 0. The zero-product principle can be applied only when a quadratic equation is in general form, with zero on one side of the equation. Solving a Quadratic Equation by Factoring 1. If necessary, rewrite the equation in the general form ax2 + bx + c = 0, moving all terms to one side, thereby obtaining zero on the other side. 2. Factor completely. 3. Apply the zero-product principle, setting each factor containing a variable equal to zero. 4. Solve the equations in step 3. 5. Check the solutions in the original equation. Solving Quadratic Equations by Factoring EXAMPLE 7 Solve by factoring: a. 4x2 - 2x = 0 b. 2x2 + 7x = 4. Solution 4x2 - 2x = 0 a. The given equation is in general form, with zero on one side. 2x12x - 12 = 0 2x = 0 or 2x - 1 = 0 x = 0 2x = 1 1 x = 2 Factor. Use the zero-product principle and set each factor equal to zero. Solve the resulting equations. Check the proposed solutions, 0 and 12 , in the original equation. Check 0: 4x2 - 2x = 0 4 # 02 - 2 # 0 ⱨ 0 0 - 0ⱨ0 0 = 0, true Check 12 : 4x2 - 2x = 0 4 A 12 B 2 - 2 A 12 B ⱨ 0 4 A 14 B - 2 A 12 B ⱨ 0 1 - 1ⱨ0 0 = 0, The solution set is E 0, 12 F . b. 2x2 + 7x = 4 2 2x + 7x - 4 = 4 - 4 This is the given equation. Subtract 4 from both sides and write the quadratic equation in general form. 2x2 + 7x - 4 = 0 12x - 121x + 42 = 0 2x - 1 = 0 or x + 4 = 0 2x = 1 x = 12 true x = -4 Simplify. Factor. Use the zero-product principle and set each factor equal to zero. Solve the resulting equations. Section P.7 Equations 91 Check the proposed solutions, 12 and -4, in the original equation. Check 12 : 2x + 7x = 4 1 2 2A B + 7A1B ⱨ 4 Check -4: 2x2 + 7x = 4 21 - 422 + 71 -42 ⱨ 4 2 2 2 1 2 + 32 + 1 - 282 ⱨ 4 4 = 4, true 7 2 ⱨ4 4 = 4, true The solution set is E -4, 12 F . Check Point 7 Solve by factoring: a. 3x2 - 9x = 0 � Solve quadratic equations by the square root property. b. 2x2 + x = 1. Quadratic Equations and the Square Root Property Quadratic equations of the form u2 = d, where u is an algebraic expression and d is a nonzero real number, can be solved by the square root property. First, isolate the squared expression u2 on one side of the equation and the number d on the other side. Then take the square root of both sides. Remember, there are two numbers whose square is d. One number is 2d and one is - 2d. We can use factoring to verify that u2 = d has these two solutions. u2 = d u2 - d = 0 This is the given equation. Move all terms to one side and obtain zero on the other side. A u + 2d B A u - 2d B = 0 Factor. u + 2d = 0 or u - 2d = 0 u = 2d u = - 2d Set each factor equal to zero. Solve the resulting equations. Because the solutions differ only in sign, we can write them in abbreviated notation as u = ; 2d. We read this as “u equals positive or negative square root of d” or “u equals plus or minus square root of d.” Now that we have verified these solutions, we can solve u2 = d directly by taking square roots. This process is called the square root property. The Square Root Property If u is an algebraic expression and d is a positive real number, then u2 = d has exactly two solutions: If u2 = d, then u = 2d or u = - 2d. Equivalently, EXAMPLE 8 If u2 = d, then u = ; 2d. Solving Quadratic Equations by the Square Root Property Solve by the square root property: a. 3x2 - 15 = 0 b. 1x - 222 = 6. Solution To apply the square root property, we need a squared expression by itself on one side of the equation. 3x2-15=0 We want x2 by itself. (x-2)2=6 The squared expression is by itself. 92 Chapter P Prerequisites: Fundamental Concepts of Algebra a. 3x2 - 15 3x2 x2 x = 25 or x = = = = 0 15 5 - 25 This is the original equation. Add 15 to both sides. Divide both sides by 3. Apply the square root property. Equivalently, x = ; 25. By checking both proposed solutions in the original equation, we can confirm that the solution set is E - 25, 25 F or E ; 25 F . b. 1x - 222 = 6 x - 2 = ; 26 x = 2 ; 26 This is the original equation. Apply the square root property. Add 2 to both sides. By checking both values in the original equation, we can confirm that the solution set is E 2 + 26, 2 - 26 F or E 2 ; 26 F . Check Point 8 Solve by the square root property: b. 1x + 522 = 11. a. 3x2 - 21 = 0 � Solve quadratic equations by completing the square. Quadratic Equations and Completing the Square How do we solve an equation in the form ax2 + bx + c = 0 if the trinomial ax2 + bx + c cannot be factored? We cannot use the zero-product principle in such a case. However, we can convert the equation into an equivalent equation that can be solved using the square root property.This is accomplished by completing the square. Completing the Square b 2 If x2 + bx is a binomial, then by adding a b , which is the square of half the 2 coefficient of x, a perfect square trinomial will result. That is, b 2 b 2 x2 + bx + a b = a x + b . 2 2 We can solve any quadratic equation by completing the square. If the coefficient of the x2-term is one, we add the square of half the coefficient of x to both sides of the equation. When you add a constant term to one side of the equation to complete the square, be certain to add the same constant to the other side of the equation. These ideas are illustrated in Example 9. EXAMPLE 9 Solving a Quadratic Equation by Completing the Square Solve by completing the square: x2 - 6x + 4 = 0. Solution We begin by subtracting 4 from both sides. This is done to isolate the binomial x2 - 6x so that we can complete the square. x2 - 6x + 4 = 0 x2 - 6x = - 4 2 This is the original equation. Subtract 4 from both sides. Next, we work with x - 6x = - 4 and complete the square. Find half the coefficient of the x-term and square it. The coefficient of the x-term is - 6. Half of - 6 is - 3 and 1-322 = 9. Thus, we add 9 to both sides of the equation. Section P.7 Equations 93 x2 - 6x + 9 = - 4 + 9 1x - 322 = 5 x - 3 = 25 x = 3 + 25 or Add 9 to both sides of x2 - 6x = - 4 to complete the square. x - 3 = - 25 x = 3 - 25 Factor and simplify. Apply the square root property. Add 3 to both sides in each equation. The solutions are 3 ; 25 and the solution set is E 3 + 25, 3 - 25 F , or E 3 ; 25 F . Check Point � Solve quadratic equations using the quadratic formula. Deriving the Quadratic Formula 9 Solve by completing the square: x2 + 4x - 1 = 0. Quadratic Equations and the Quadratic Formula We can use the method of completing the square to derive a formula that can be used to solve all quadratic equations. The derivation given here also shows a particular quadratic equation, 3x2 - 2x - 4 = 0, to specifically illustrate each of the steps. Notice that if the coefficient of the x2-term in a quadratic equation is not one, you must divide each side of the equation by this coefficient before completing the square. General Form of a Quadratic Equation Comment A Specific Example ax2 + bx + c = 0, a 7 0 This is the given equation. 3x2 - 2x - 4 = 0 x2 + b c = 0 x + a a Divide both sides by a so that the coefficient of x2 is 1. x2 + b c x = a a b c b 2 b 2 x2+ x + a b = - + a b a 2a a 2a x2 - 2 4 x - = 0 3 3 c Isolate the binomial by adding - on both sides of a the equation. x2 - 4 2 x = 3 3 Complete the square. Add the square of half the coefficient of x to both sides. 1 2 4 1 2 2 x2- x + a- b = + a- b 3 3 3 3 (half)2 b2 b b2 c + x2 + x + = a a 4a2 4a2 ax + c 4a b 2 b2 b = - # + a 4a 2a 4a2 -4ac + b2 b 2 b = 2a 4a2 b 2 b2 - 4ac ax + b = 2a 4a2 ax + x + x + x = x = b b2 - 4ac = ; 2a B 4a2 b 2b2 - 4ac = ; 2a 2a -b 3b2 - 4ac ; 2a 2a - b ; 3b2 - 4ac 2a (half)2 2 1 4 1 x2 - x + = + 3 9 3 9 4 3 1 2 1 b = # + 3 3 3 9 Factor on the left side and obtain a common denominator on the right side. ax - Add fractions on the right side. 1 2 12 + 1 b = 3 9 1 2 13 ax - b = 3 9 Apply the square root property. Take the square root of the quotient, simplifying the denominator. Solve for x by subtracting b from both sides. 2a Combine fractions on the right side. ax - x - x - x = x = 1 13 = ; 3 A9 1 213 = ; 3 3 1 213 ; 3 3 1 ; 213 3 The formula shown at the bottom of the left column is called the quadratic formula. A similar proof shows that the same formula can be used to solve quadratic equations if a, the coefficient of the x2-term, is negative. 94 Chapter P Prerequisites: Fundamental Concepts of Algebra The Quadratic Formula The solutions of a quadratic equation in general form ax2 + bx + c = 0, with a Z 0, are given by the quadratic formula x= –b_兹b2-4ac . 2a x equals negative b plus or minus the square root of b2 − 4ac, all divided by 2a. To use the quadratic formula, write the quadratic equation in general form if necessary. Then determine the numerical values for a (the coefficient of the x2-term), b (the coefficient of the x-term), and c (the constant term). Substitute the values of a, b, and c into the quadratic formula and evaluate the expression. The ; sign indicates that there are two solutions of the equation. EXAMPLE 10 Solving a Quadratic Equation Using the Quadratic Formula Solve using the quadratic formula: 2x2 - 6x + 1 = 0. Solution The given equation is in general form. Begin by identifying the values for a, b, and c. 2x2-6x+1=0 a=2 b = −6 c=1 Substituting these values into the quadratic formula and simplifying gives the equation’s solutions. x = = = Study Tip Checking irrational solutions can be time-consuming. The solutions given by the quadratic formula are always correct, unless you have made a careless error. Checking for computational errors or errors in simplification is sufficient. = = = -b ; 3b2 - 4ac 2a Use the quadratic formula. - 1- 62 ; 41- 622 - 4122112 2#2 6 ; 236 - 8 4 6 ; 228 4 - 1- 62 = 6, 1 - 622 = 1- 621 -62 = 36, and 4122112 = 8. Complete the subtraction under the radical. 6 ; 2 27 4 228 = 24 # 7 = 2427 = 2 27 2 A 3 ; 27 B Factor out 2 from the numerator. 4 3 ; 27 = 2 Divide the numerator and denominator by 2. The solution set is b Check Point Substitute the values for a, b, and c: a = 2, b = - 6, and c = 1. 10 3 + 27 3 - 27 3 ; 27 , r or b r. 2 2 2 Solve using the quadratic formula: 2x2 + 2x - 1 = 0. Section P.7 Equations 95 쐅 Use the discriminant to determine the number and type of solutions of quadratic equations. Quadratic Equations and the Discriminant The quantity b2 - 4ac, which appears under the radical sign in the quadratic formula, is called the discriminant. Table P.4 shows how the discriminant of the quadratic equation ax2 + bx + c = 0 determines the number and type of solutions. Table P.4 The Discriminant and the Kinds of Solutions to ax2 ⴙ bx ⴙ c ⴝ 0 Discriminant b2 ⴚ 4ac Kinds of Solutions to ax2 ⴙ bx ⴙ c ⴝ 0 b2 - 4ac 7 0 Two unequal real solutions: If a, b, and c are rational numbers and the discriminant is a perfect square, the solutions are rational. If the discriminant is not a perfect square, the solutions are irrational. b2 - 4ac = 0 One solution (a repeated solution) that is a real number: If a, b, and c are rational numbers, the repeated solution is also a rational number. b2 - 4ac 6 0 No real solutions EXAMPLE 11 Using the Discriminant Compute the discriminant of 4x2 - 8x + 1 = 0. What does the discriminant indicate about the number and type of solutions? Solution Begin by identifying the values for a, b, and c. 2x2-6x+1=0 a=2 b = −6 c=1 Substitute and compute the discriminant: b2 - 4ac = 1-822 - 4 # 4 # 1 = 64 - 16 = 48. The discriminant is 48. Because the discriminant is positive, the equation 4x2 - 8x + 1 = 0 has two unequal real solutions. 11 Compute the discriminant of 3x2 - 2x + 5 = 0. What does the discriminant indicate about the number and type of solutions? Check Point 쐈 Determine the most efficient method to use when solving a quadratic equation. Determining Which Method to Use All quadratic equations can be solved by the quadratic formula. However, if an equation is in the form u2 = d, such as x2 = 5 or 12x + 322 = 8, it is faster to use the square root property, taking the square root of both sides. If the equation is not in the form u2 = d, write the quadratic equation in general form 1ax2 + bx + c = 02. Try to solve the equation by factoring. If ax2 + bx + c cannot be factored, then solve the quadratic equation by the quadratic formula. Because we used the method of completing the square to derive the quadratic formula, we no longer need it for solving quadratic equations. However, we will use completing the square later in the book to help graph circles and other kinds of equations. Table P.5 on the next page summarizes our observations about which technique to use when solving a quadratic equation. 96 Chapter P Prerequisites: Fundamental Concepts of Algebra Table P.5 Determining the Most Efficient Technique to Use When Solving a Quadratic Equation Description and Form of the Quadratic Equation Most Efficient Solution Method ax2 + bx + c = 0 and ax2 + bx + c can be factored easily. Factor and use the zero-product principle. 3x2 + 5x - 2 = 0 13x - 121x + 22 = 0 or x + 2 = 0 3x - 1 = 0 1 x = x = -2 3 ax2 + bx = 0 The quadratic equation has no constant term. 1c = 02 Factor and use the zero-product principle. 6x2 + 9x = 0 3x12x + 32 = 0 or 2x + 3 = 0 3x = 0 2x = - 3 x = 0 x = - 23 ax2 + c = 0 The quadratic equation has no x-term. 1b = 02 Solve for x2 and apply the square root property. u2 = d; u is a first-degree polynomial. Example 7x2 - 4 = 0 7x2 = 4 4 x2 = 7 x = ; 4 A7 = ; 27 2 = ; # 27 27 27 2 = ; 227 7 1x + 422 = 5 Use the square root property. x + 4 = ; 25 x = - 4 ; 25 ax2 + bx + c = 0 and ax2 + bx + c cannot be factored or the factoring is too difficult. Use the quadratic formula: x¤-2x-6=0 -b ; 3b - 4ac . 2a 2 x = a=1 x = = b = −2 c = −6 -1-22 ; 41- 222 - 41121 - 62 2(1) 2 ; 44 + 24 2(1) 2 ; 228 2 ; = 2 2A1 2 ; 227 = = 2 = = 1 ; 27 쐉 Solve radical equations. 24 27 2 ; 27 B 2 Radical Equations A radical equation is an equation in which the variable occurs in a square root, cube root, or any higher root. An example of a radical equation is 1x = 9. We solve the equation by squaring both sides: Squaring both sides eliminates the square root. A兹xB =92 2 x=81. The proposed solution, 81, can be checked in the original equation, 1x = 9. Because 281 = 9, the solution is 81 and the solution set is 5816. Section P.7 Equations 97 In general, we solve radical equations with square roots by squaring both sides of the equation. We solve radical equations with nth roots by raising both sides of the equation to the nth power. Unfortunately, if n is even, all the solutions of the equation raised to the even power may not be solutions of the original equation. Consider, for example, the equation x = 4. If we square both sides, we obtain x2 = 16. Solving this equation using the square root property, we obtain x = ; 216 = ; 4. The new equation x = 16 has two solutions, - 4 and 4. By contrast, only 4 is a solution of the original equation, x = 4. For this reason, when raising both sides of an equation to an even power, always check proposed solutions in the original equation. Here is a general method for solving radical equations with nth roots: 2 Solving Radical Equations Containing nth Roots 1. If necessary, arrange terms so that one radical is isolated on one side of the equation. 2. Raise both sides of the equation to the nth power to eliminate the isolated nth root. 3. Solve the resulting equation. If this equation still contains radicals, repeat steps 1 and 2. 4. Check all proposed solutions in the original equation. Extra solutions may be introduced when you raise both sides of a radical equation to an even power. Such solutions, which are not solutions of the given equation, are called extraneous solutions or extraneous roots. EXAMPLE 12 Solving a Radical Equation Solve: 22x - 1 + 2 = x. Solution Step 1 Isolate a radical on one side. We isolate the radical, 22x - 1, by subtracting 2 from both sides. 22x - 1 + 2 = x 22x - 1 = x - 2 Study Tip Be sure to square both sides of an equation. Do not square each term. Correct: A 22x - 1 B 2 = 1x - 222 Incorrect! A 22x - 1 B 2 = x2 - 2 2 This is the given equation. Subtract 2 from both sides. Step 2 Raise both sides to the nth power. Because n, the index, is 2, we square both sides. A 22x - 1 B 2 = 1x - 222 2x - 1 = x2 - 4x + 4 Simplify. Use the formula 1A - B22 = A2 - 2AB + B2 on the right side. Step 3 Solve the resulting equation. Because of the x2-term, the resulting equation is a quadratic equation. We can obtain 0 on the left side by subtracting 2x and adding 1 on both sides. 2x - 1 = x2 - 4x + 4 The resulting equation is quadratic. 0 = x2 - 6x + 5 0 = 1x - 121x - 52 x - 1 = 0 or x - 5 = 0 x = 1 x = 5 Write in general form, subtracting 2x and adding 1 on both sides. Factor. Set each factor equal to 0. Solve the resulting equations. 98 Chapter P Prerequisites: Fundamental Concepts of Algebra Step 4 Check the proposed solutions in the original equation. Check 1: Check 5: 22x - 1 + 2 = x 22x - 1 + 2 = x 22 # 1 - 1 + 2 ⱨ 1 22 # 5 - 1 + 2 ⱨ 5 21 + 2 ⱨ 1 1 + 2ⱨ1 29 + 2 ⱨ 5 3 + 2ⱨ5 3 = 1, 5 = 5, false true Thus, 1 is an extraneous solution.The only solution is 5, and the solution set is 556. Check Point 12 Solve: 2x + 3 + 3 = x. Exercise Set P.7 Practice Exercises 25. 5 6 1 = 2 x - 4 x + 2 x - 2x - 8 26. 2 8 1 = 2 x - 3 x + 1 x - 2x - 3 In Exercises 1–16, solve each linear equation. 1. 7x - 5 = 72 2. 6x - 3 = 63 3. 11x - 16x - 52 = 40 4. 5x - 12x - 102 = 35 5. 2x - 7 = 6 + x 6. 3x + 5 = 2x + 13 7. 7x + 4 = x + 16 8. 13x + 14 = 12x - 5 9. 31x - 22 + 7 = 21x + 52 x + 1 1 2 - x = + 4 6 3 11. x + 3 3 x - 5 = + 6 8 4 12. 13. x x - 3 = 2 + 4 3 14. 5 + x - 2 x + 3 = 3 8 x + 1 x + 2 x - 3 x + 2 3x = 5 = 16. 3 7 5 2 3 Exercises 17–26 contain rational equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 15. 19. 8x 8 = 4 x + 1 x + 1 21. 3 1 2 + = 2x - 2 2 x - 1 22. 23. 3 -4 - 7 = 18. x + 4 x + 4 20. 27. I = Prt for P 28. C = 2pr for r 29. T = D + pm for p 30. P = C + MC for M 1 2 h1a 32. A = 21 h1a + b2 for b 31. A = 10. 21x - 12 + 3 = x - 31x + 12 1 11 + 5 = 17. x - 1 x - 1 In Exercises 27–42, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? 2 x = - 2 x - 2 x - 2 + b2 for a 33. S = P + Prt for r F 35. B = for S S - V 34. S = P + Prt for t C 36. S = for r 1 - r 37. IR + Ir = E for I 1 1 1 39. + = for f p q f f1f2 41. f = for f1 f1 + f2 38. A = 2lw + 2lh + 2wh for h 1 1 1 40. for R1 = + R R1 R2 f1f2 42. f = for f2 f1 + f2 In Exercises 43–54, solve each absolute value equation or indicate the equation has no solution. 43. ƒ x - 2 ƒ = 7 44. ƒ x + 1 ƒ = 5 45. ƒ 2x - 1 ƒ = 5 46. ƒ 2x - 3 ƒ = 11 47. 2 ƒ 3x - 2 ƒ = 14 48. 3 ƒ 2x - 1 ƒ = 21 49. 2 ` 4 - 50. 4 ` 1 - 5 x ` + 6 = 18 2 3 x ` + 7 = 10 4 51. ƒ x + 1 ƒ + 5 = 3 52. ƒ x + 1 ƒ + 6 = 2 5 1 3 = + x + 3 2x + 6 x - 2 53. ƒ 2x - 1 ƒ + 3 = 3 54. ƒ 3x - 2 ƒ + 4 = 4 1 2x 2 = 2 x + 1 x - 1 x - 1 55. x2 - 3x - 10 = 0 56. x2 - 13x + 36 = 0 57. x2 = 8x - 15 58. x2 = - 11x - 10 59. 5x2 = 20x 60. 3x2 = 12x 2 32 4 + = 2 24. x + 5 x - 5 x - 25 In Exercises 55–60, solve each quadratic equation by factoring. Section P.7 Equations 99 In Exercises 61–66, solve each quadratic equation by the square root property. 61. 3x2 = 27 62. 5x2 = 45 63. 5x2 + 1 = 51 64. 3x2 - 1 = 47 65. 31x - 422 = 15 66. 31x + 422 = 21 In Exercises 67–74, solve each quadratic equation by completing the square. 2 67. x + 6x = 7 2 69. x - 2x = 2 2 2 68. x + 6x = - 8 2 70. x + 4x = 12 2 71. x - 6x - 11 = 0 72. x - 2x - 5 = 0 73. x2 + 4x + 1 = 0 74. x2 + 6x - 5 = 0 In Exercises 75–82, solve each quadratic equation using the quadratic formula. 2 75. x + 8x + 15 = 0 2 77. x + 5x + 3 = 0 2 76. x + 8x + 12 = 0 2 78. x + 5x + 2 = 0 2 80. 5x + x - 2 = 0 2 82. 3x2 = 6x - 1 79. 3x - 3x - 4 = 0 81. 4x = 2x + 7 2 Compute the discriminant of each equation in Exercises 83–90. What does the discriminant indicate about the number and type of solutions? 83. x2 - 4x - 5 = 0 84. 4x2 - 2x + 3 = 0 85. 2x2 - 11x + 3 = 0 86. 2x2 + 11x - 6 = 0 2 2 87. x = 2x - 1 88. 3x = 2x - 1 89. x2 - 3x - 7 = 0 90. 3x2 + 4x - 2 = 0 In Exercises 91–114, solve each quadratic equation by the method of your choice. 2 91. 2x - x = 1 2 92. 3x - 4x = 4 2 94. 5x = 6 - 13x 2 95. 3x = 60 96. 2x2 = 250 97. x2 - 2x = 1 98. 2x2 + 3x = 1 93. 5x + 2 = 11x 99. 12x + 321x + 42 = 1 101. 13x - 42 = 16 2 2 2 100. 12x - 521x + 12 = 2 117. 2x + 3 = x - 3 118. 2x + 10 = x - 2 121. x - 22x + 5 = 5 122. x - 2x + 11 = 1 119. 22x + 13 = x + 7 123. 22x + 19 - 8 = x In Exercises 125–134, solve each equation. 125. 25 - 32 + 5x - 31x + 224 = -312x - 52 - 351x - 12 - 3x + 34 126. 45 - 34 - 2x - 41x + 724 = -411 + 3x2 - 34 - 31x + 22 - 212x - 524 127. 7 - 7x = 13x + 221x - 12 128. 10x - 1 = 12x + 122 129. ƒ x2 + 2x - 36 ƒ = 12 130. ƒ x2 + 6x + 1 ƒ = 8 1 1 5 131. 2 = + 2 x + 2 x - 3x + 2 x - 4 x 1 x - 1 132. + = 2 x - 2 x - 3 x - 5x + 6 133. 2x + 8 - 2x - 4 = 2 134. 2x + 5 - 2x - 3 = 2 In Exercises 135–136, list all numbers that must be excluded from the domain of each rational expression. 7 3 135. 136. 2x2 + 4x - 9 2x2 - 8x + 5 Application Exercises The latest guidelines, which apply to both men and women, give healthy weight ranges, rather than specific weights, for your height. The further you are above the upper limit of your range, the greater are the risks of developing weight-related health problems. The bar graph shows these ranges for various heights for people between the ages of 19 and 34, inclusive. 102. 12x + 722 = 25 Healthy Weight Ranges for Men and Women, Ages 19 to 34 2 104. 9 - 6x + x = 0 105. 4x2 - 16 = 0 106. 3x2 - 27 = 0 220 107. x2 = 4x - 2 108. x2 = 6x - 7 200 109. 2x2 - 7x = 0 110. 2x2 + 5x = 3 180 112. 1 1 1 = + x x + 3 4 6 28 2x + = - 2 x - 3 x + 3 x - 9 3 5 x2 - 20 114. + = 2 x - 3 x - 4 x - 7x + 12 113. In Exercises 115–124, solve each radical equation. Check all proposed solutions. 115. 23x + 18 = x 116. 220 - 8x = x Lower end of range Weight (pounds) 1 1 1 = + x x + 2 3 124. 22x + 15 - 6 = x Practice Plus 103. 3x - 12x + 12 = 0 111. 120. 26x + 1 = x - 1 160 146 120 111 118 125 132 195 184 174 164 155 140 Upper end of range 140 148 100 80 60 5⬘4⬙ 5⬘6⬙ 5⬘8⬙ 5⬘10⬙ Height Source: U.S. Department of Health and Human Services 6⬘0⬙ 6⬘2⬙ 100 Chapter P Prerequisites: Fundamental Concepts of Algebra The mathematical model Use the formula at the bottom of the previous column to solve Exercises 141–142. How many liters of pure peroxide should be added to produce a new product that is 28% peroxide? 140. Suppose that x liters of pure acid are added to 200 liters of a 35% acid solution. a. Write a formula that gives the concentration, C, of the new mixture. (Hint: See Exercise 139.) b. How many liters of pure acid should be added to produce a new mixture that is 74% acid? A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died. Fatal Crashes per 100 Million Miles Driven Age of U.S. Drivers and Fatal Crashes 18 17.7 16.3 15 12 9.5 9 8.0 6.2 6 4.1 3 0 16 18 20 25 2.8 2.4 3.0 35 45 55 Age of Drivers 3.8 65 75 79 Source: Insurance Institute for Highway Safety The number of fatal vehicle crashes per 100 million miles, y, for drivers of age x can be modeled by the formula y = 0.013x2 - 1.19x + 28.24. Average Nonprogram Minutes in an Hour of Prime-Time Cable TV 16 15 14.3 14 13 14.6 13.5 12.5 12 11 10 2005 139. A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure peroxide are added to 500 liters of its 10% solution, the concentration, C, of the new mixture is given by x + 0.115002 C = . x + 500 By 2005, the amount of “clutter,” including commercials and plugs for other shows, had increased to the point where an “hour-long” drama on cable TV was 45.4 minutes. The bar graph shows the average number of nonprogram minutes in an hour of prime-time cable television. 2002 138. Use the formula to find a healthy weight for a person whose height is 6¿0–. (Hint: H = 12 because this person’s height is 12 inches over 5 feet.) How many pounds is this healthy weight below the upper end of the range shown by the bar graph on the previous page? 142. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown by the bar graph in the previous column? 1999 137. Use the formula to find a healthy weight for a person whose height is 5¿6–. (Hint: H = 6 because this person’s height is 6 inches over 5 feet.) How many pounds is this healthy weight below the upper end of the range shown by the bar graph on the previous page? 1996 describes a weight, W, in pounds, that lies within the healthy weight range shown by the bar graph on the previous page for a person whose height is H inches over 5 feet. Use this information to solve Exercises 137–138. 141. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown by the bar graph in the previous column? Non-Program Minutes in an Hour W - 3H = 53 2 Year Source: Nielsen Monitor-Plus M = 0.7 1x + 12.5, The data can be modeled by the formula where M is the average number of nonprogram minutes in an hour of prime-time cable x years after 1996. Use the formula to solve Exercises 143–144. 143. Assuming the trend from 1996 through 2005 continues, use the model to project when there will be 15.1 cluttered minutes in every prime-time cable TV hour. 144. Assuming the trend from 1996 through 2005 continues, use the model to project when there will be 16 cluttered minutes in every prime-time cable TV hour. Writing in Mathematics 145. What is a linear equation in one variable? Give an example of this type of equation. 146. Explain how to determine the restrictions on the variable for the equation 3 4 7 . + = 2 x + 5 x - 2 x + 3x - 6 147. What does it mean to solve a formula for a variable? 148. Explain how to solve an equation involving absolute value. 149. Why does the procedure that you explained in Exercise 148 not apply to the equation ƒ x - 2 ƒ = - 3? What is the solution set for this equation? 150. What is a quadratic equation? 151. Explain how to solve x2 + 6x + 8 = 0 using factoring and the zero-product principle. 152. Explain how to solve x2 + 6x + 8 = 0 by completing the square. 153. Explain how to solve x2 + 6x + 8 = 0 using the quadratic formula. 154. How is the quadratic formula derived? 155. What is the discriminant and what information does it provide about a quadratic equation? Section P.8 Modeling with Equations 101 156. If you are given a quadratic equation, how do you determine which method to use to solve it? 157. In solving 22x - 1 + 2 = x, why is it a good idea to isolate the radical term? What if we don’t do this and simply square each side? Describe what happens. 158. What is an extraneous solution to a radical equation? Critical Thinking Exercises Make Sense? In Exercises 159–162, determine whether each statement makes sense or does not make sense, and explain your reasoning. 159. The model P = - 0.18n + 2.1 describes the number of pay phones, P, in millions, n years after 2000, so I have to solve a linear equation to determine the number of pay phones in 2006. 160. Although I can solve 3x + 15 = 41 by first subtracting 51 from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator. 161. Because I want to solve 25x2 - 169 = 0 fairly quickly, I’ll use the quadratic formula. 162. When checking a radical equation’s proposed solution, I can substitute into the original equation or any equation that is part of the solution process. In Exercises 163–166, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 163. The equation 12x - 322 = 25 is equivalent to 2x - 3 = 5. Section P.8 164. Every quadratic equation has two distinct numbers in its solution set. 165. The equations 3y - 1 = 11 and 3y - 7 = 5 are equivalent. 166. The equation ax2 + c = 0, a Z 0, cannot be solved by the quadratic formula. 7x + 4 167. Find b such that + 13 = x will have a solution set b given by 5 - 66. 168. Write a quadratic equation in general form whose solution set is 5 - 3, 56. C - S N. 169. Solve for C: V = C L 2 170. Solve for t: s = - 16t + v0t. Preview Exercises Exercises 171–173 will help you prepare for the material covered in the next section. 171. Jane’s salary exceeds Jim’s by $150 per week. If x represents Jim’s weekly salary, write an algebraic expression that models Jane’s weekly salary. 172. A long-distance telephone plan has a monthly fee of $20 with a charge of $0.05 per minute for all long-distance calls. Write an algebraic expression that models the plan’s monthly cost for x minutes of long-distance calls. 173. If the width of a rectangle is represented by x and the length is represented by x + 200, write a simplified algebraic expression that models the rectangle’s perimeter. Modeling with Equations Objective How Long It Takes to Earn $1000 � Use equations to solve problems. Howard Stern Radio host 24 sec. Dr. Phil McGraw Television host 2 min. 24 sec. Brad Pitt Actor 4 min. 48 sec. Kobe Bryant Basketball player 5 min. 30 sec. Chief executive U.S. average 2 hr. 55 min. Doctor, G.P. U.S. average 13 hr. 5 min. High school teacher U.S. average 43 hours Janitor U.S. average 103 hours Source: Time I n this section, you’ll see examples and exercises focused on how much money Americans earn. These situations illustrate a step-by-step strategy for solving problems. As you become familiar with this strategy, you will learn to solve a wide variety of problems.